IN   MEMORIAM 

FLORIAN  CAJOF 


ELEMENTARY 


SYNTHETIC    GEOMETRY. 


BY 


GEORGE  BRUCE  HALSTED, 

A.B.^  A.M.,  and  ex-Fellow  of  Princeton  College  ;   Ph.D.  and  ex-Fellow  of  Johns  Hopkin 
University ;  Professor  of  Mathematics  in  the  University  of  Texas. 


NEW   YORK: 

JOHN    WILEY    &    SONS, 

53  East  Tenth  Street, 

1892. 


Copyright,  iSgz, 

BV 

GEORGE  BRUCE  HALSTED. 


CAJORI 


]FEMU8  B»03., 

ROBERT  Dromhont>,  Pnnters, 

Eleotrotvper,  ^^^  g^.'-eets 

tU&m  Pearl  Street.  *    ^^^  ^^^^ 
New  York. 


PREFACE. 


My  conception  of  Comparative  Geometry  dates  back  to 
1877,  when,  having  had  the  good  fortune  in  BerHn  to  meet  a 
copy  of  Lobatschewsky,  then  rare  and  unappreciated,  I  was 
enjoying  a  comparison  of  it  with  my  beloved  EucHd. 

I  have  at  last  mustered  courage  to  put  my  Pure  Spherics 
where  it  belongs.  But  as  my  first  book  is  geometry  with  one 
parallel  geodesic,  my  second  book  geometry  with  no  parallel 
geodesic,  my  third  book  should  be  geometry  with  more  than 
one  parallel  geodesic  through  the  same  point  (the  Lobat- 
schewsky-Bolyai  elementary  geometry).  So  it  shall  be,  if, 
after  another  decade,  I  write  still  another  geometry. 

George  Bruce  Halsted. 

2407  Guadalupe  Street,  Austin,  Texas. 

ill 


918190 


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in  2008  witin  funding  from 

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CONTENTS. 


BOOK   I. 

PAGB 

Symmetry,  Symcentry,  and  Congruence, i 

CHAPTER  SKCTIONS 

I.  The  Primary  Concepts  of  Geometry,    .    .    .       1-54 
II.  The  Circle, 55-io9 

III.  The  Fundamental  Problems 110-119 

IV.  Symmetry  and  Symcentry 120-152 

V.  Tangents, 1 53-171 

VI.  Chords 172-183 

VII.  Two  Circles 184-186 

VIII.  Parallels 187-205 

IX.  The  Triangle 206-227 

X.   Polygons, 228-243 

XI.   Periphery  Angles, 244-261 

XII.  The  Symmetrical  Triangle, 262-268 

XIII.  The  Symcentral  Quadrilaterals, 269-280 

XIV.  Symmetrical  Quadrilateral, 281-299 

XV.  Congruence  of  Triangles, 300-313 

BOOK   II. 
Pure  Spherics 60 

CHAPTER  SECTIONS 

I.  Primary  Concepts, 314-340 

II.  Symcentry  on  the  Sphere 34^-343 

III.  Symmetr)'  on  the  Sphere, 344-383 

IV.  The  Symcentral  Quadrilateral :  384-389 

•V.  Spherical  Triangles, 390-428 


VI  CONTENTS. 


BOOK    III. 


PAGE 

Equivalence 85 


BOOK   IV. 
Proportion 92 

BOOK  V. 
Similarity 95 

BOOK  VI. 
Mensuration 104. 

BOOK  VII. 
Modern  Geometry, 117- 

CHAPTER  SECTIONS 

I.  Transversals 522-533 

II.  Harmonic  Ranges  and  Pencils 534-55° 

III.  Principle  of  Duality 551-564 

IV.  Complete  Quadrilateral  and  Quadrangle,     .   565-572 
V.  Inversion 573-581 

VI,   Pole  and  Polar  with  respect  to  a  Circle,  .     .  582-599 
VII.  Cross  Ratio 600-623 

BOOK  VIII. 
Recent  Geometry 146 

CHAPTER  SECTIONS 

I.   Anti-parallels,  Isogonals,  Symmedians,     .     .  624-660 
II.  The  Brocard  Points 661-675 


TABLE   OF    SYMBOLS. 

bi' bisector. 

e.g exempli  gratia  [for  example]. 

p't point. 

p'ts points. 

quad' quadrilateral. 

r't right. 

r't  bi' right  bisector  [perpendicular  bisector]. 

sq' square. 

s't straight. 

s'ts straights. 

O circle. 

• surface  of  circle. 

O* circles. 

A. triangle. 

A" triangles. 

A spherical  triangle. 

.• therefore. 

~ similar. 

~C center  of  similitude. 

= equivalent. 

^ congruent. 

symcentral. 

•|' symmetrical. 

II parallel. 

IP parallels. 

I!  g'm parallelogram. 

J_ perpendicular. 

J_^ perpendiculars. 

-f- plus. 

— minus. 

< less  than. 

> greater  than. 

^ angle. 

^ABC.  .  .  .angle  from  ray  BA  to  ray  BC. 

'^ab angle  from  ray  a  to  ray  b. 

QC{r) circle  with  center  C  and  radius  r. 

A perspective. 

A\C  .  : center  of  perspective. 

A projective. 


ELEMENTARY  SYNTHETIC   GEOMETRY. 


BOOK    I. 

SYMMETRY,  SYMCENTRY,  AND  CONGRUENCE. 


CHAPTER    I. 


THE   PRIMARY   CONCEPTS   OF  GEOMETRY. 


1.  A  natural  object,  say  a  crystal,  is  bounded  ;  and  this 
boundary  divides  it  from  the  air  around  it,  but  is  not 
a  thin  film  of  the  crystal  itself.  It  is  where  the 
crystal  ends  and  the  air  begins.  It  is  also  a 
boundary  of  the  air  where  it  joins  the  crystal,  but 
it  is  not  air.  It  is  the  boundary  between  the  two, 
and  is  common  to  the  crystal  and  the  air. 

2.  A  boundary    of   the  sort  capable    of    wholly 
enclosing  a  solid,    so    that    nothing  could  get  into 
the  solid  except  through  this  boundary,  but  itself  no  solid,  is 
called  a  S7irface. 

3.  Surface  is  an  ideal  or  imaginary  concept  drawn  from  the 
apparent  (not  real)  boundaries  of  physical  objects.  We  natu- 
rally associate  the  surface  with  the  limited  solid,  not  with  the 


Fig. 


2  .^LEMBA'tA^HY  SYNTHETIC  GEOMETRY. 

surrounding  air.  Thus  we  think  of  the  colored  surface  of  the 
crystal  as  belonging  to  the  crystal ;  and  if  yellow  oil  lies  on 
the  water  in  a  glass,  we  think  of  the  under  surface  of  the  oil 
as  yellow  and  belonging  only  to  the  oil :  while  a  mathematical 
surface  pertains  equally  to  the  two  solids  that  it  separates. 

4.  These  ideal  mathematical  surfaces  may  be  dealt  with  as 
existing  by  themselves,  and  as  movable.  In  illustration  of  this, 
think  of  a  shadow. 

5.  A  surface  may  be  finite  yet  unbounded  in  the  sense  of 
having  no  abrupt  or  natural  stopping-place  on  it,  no  visible 
break  or  obvious  limit  in  it.  Such  is  the  surface  of  an  egg. 
Set  it  up  in  an  egg-cup,  and  run  a  pencil-mark  around  it. 
Then  you  may  think  of  the  surface  of  the  egg  as  divided  into 
three  parts,  the  white  surface  within  the  cup,  the  ribbon-like 
black  surface  of  the  pencil-mark,  and  the  white  surface  above 
this  black  ribbon. 

6.  Between  the  black  surface  and  the  two  white  surfaces 
are  two  boundaries  which  are  neither  black  nor  white.  These 
boundaries  are  not  thin  strips  of  surface  any  more  than  the 
surface  is  a  thin  layer  of  solid.  Where  a  white  surface  meets  a 
black  there  is  a  common  boundary  of  both,  dividing  each  from 
the  other,  and  belonging  to  both. 

7.  A  boundary  of  the  sort  capable  of  wholly 
enclosing  a  piece  of  surface  so  that  nothing  mov- 
ing in  the  surface  could  enter  this  piece  of  surface 
except  through  this  boundary,  but  itself  no  sur- 
face, is  called  a  line. 

Fig.  2. 

8.  A  boundary  between  two  adjacent  pieces  of  a  line,  and 
common  to  both  pieces,  but  itself  no  line,  is  called  2i point. 


9.  Two  lines  cross  or  intersect  in  a  point. 

Fig,  3. 


PRIMARY  CONCEPTS. 


10.  Two  surfaces  intersect  in  a  line. 


Fig. 


11.  When  a  chalk  mark  is  drawn  across  a  blackboard,  each 
of  the  two  edges,  neither  white  nor  black,  is  a  line. 

12.  When  one  chalk  mark  crosses  another,  four 
points  are  fixed  by  the  crossing  of  the  four  edges. 

Fig.  s. 

13.  Surfaces,  lines,  or  points,  or  any  combinations  of  them, 
are  called  figures. 

14.  Any  figure    may  be    looked   upon    as  two   coincident 
figures;     Mathematical  figures  wholly  lack  impenetrability. 

15.  If  we  imagine  a  figure  to  move,  we 
may  also  suppose  it  to  leave  behind  its 
outline  or  trace.  „     ^ 

Fig.  6. 

16.  Two  coincident  figures  cannot  be  distinguished  from 
one  another  unless  they  be  separated  by  moving  one. 

17.  Assumption  I.     Figures  may  be  moved  about,  with- 
out any  other  change. 

18.  Figures  which    can    be    made  to    coincide  are  called 
congruent. 

19.  If  a  solid  has,  as  part  of  its  boundary,  a  piece  of  sur- 
face which  appears  the  same  from  within  the  solid  as  from 
without,  and  if  any  two  of  three  such  soHds  will  fit  each  other 
all  over  these  surfaces,  then  each  of  these 
surfaces  is  called  plane.  Such  a  surface' 
unbounded  is  called  3.  plane. 

20.  Any  piece  of  a  plane  will  slide  in 
the  plane,  and  after  being  turned  over 
will  fit  the  plane. 

^  Fig.  7. 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


21.  The  intersection  of  two  planes  is  called  a  straight  line, 
or  simply  a  straight. 

22.  A  straight  is  a  line  in  a  plane  which  appears  the  same 
, from  both  the  regions  bounded  by  it  in  the 

„     „  plane. 

Fig.  8.  ^ 

(~-\s\  23.  A  piece  of  the  plane  with  part  of  the 

straight  as  one  of  its  boundaries  would  fit  all 
along  the  straight  from  both  sides. 


Fig.  9. 


24.  Assumption  II.     If  two  straights  have  two  points  in 
common  they  coincide  throughout. 

25.  A  straight  with  two  points  in 
a  plane  lies  wholly  in  that  plane. 

For  it  lies  in  a  plane,  and  if  this 
is  another  plane  the  two  intersect  in  a 
straight  which  has  two  points  in  com- 
mon with  the  given  straight. 


Fig.  10. 


26.  Assumed  Construction    I.     A   straight    can    be   drawn 

through  any  two  points. 

, ^,        27.  A  sect  is  the  piece  of  a  straight  between 

two  definite  points. 


Fig.  II. 


28.  A  curve  is  a  line  no  part  of  which  is  straight. 


Fig,  12. 


29.  Assumption  III.  A  figure  with  two  points  fixed  can 
still  be  moved,  and  the  whole  figure  partakes  of  the  motion, 
except  the  straight  through  the  two  fixed  points. 

Such  motion  is  called  revolution  about  this  straight  as  axis. 
It  may  be  continued  until  each  point  of  the  figure  coincides 
with  its  trace.  Such  a  turning  is  spoken  of  as  one  complete 
revolution,  or  simply,  a  revolution. 

30.  If  a  third  point,  not  on  the  axis,  be  fixed,  all  motion  of 
a  rigid  figure  is  prevented. 


PRIMARY  CONCEPTS. 


Fig.  13. 


Fig.  14. 


31.  Three  points  not  on  a  straight  are  necessary  and  suf- 
ficient to  determine  a  plane. 

32.  Any  straight  in  a  plane  cuts  it  into  two  parts  called 
Jiemiplanes. 

33.  By  a  half-revolution  of  their  plane  about  the  coTimon 
straight,  either  of  two  hemiplanes  may  be  brought  into  coinci- 
dence with  the  trace  of  the  other.  Thus  one  hemiplane  may 
be  thought  of  as  made  to  coincide  with  the  other  by  folding 
over  along  the  common  axis. 

34.  Any  point  in  a  straight  cuts  it  into  two    a         o b 

parts  called  rays. 

35.  The  figure  so  formed  is  a  special 
case  of  the  figure  formed  by  two  rays  going 
out  from  the  same  point,  called  a  bi-radial. 

36.  A  bi-radial  lies  wholly  in  one  plane. 

37.  One  ray,  a,  of  a  bi-radial,  may  be  brought  into  coin- 
cidence with  the  other  ray,  b,  by  a  turning  in  the  plane,  or 
rotation,  about  the  common  point,  or  vertex  O ;  and  this 
turning  may  be  in  the  sense  indicated  by  the  arrow  in  Fig.  14, 
or  in  the  opposite  sense. 

38.  To  fix  that  sense  of  rotation  which  is  to  be  considered 
as  positive  (which  kind  is  meant  if  nothing  else  is  stated),  we 
take  the  turning  of  a  ray  in  the  sense  opposite  to  that  of  the 
hands  of  a  watch  as  positive.  The  watch  hands,  then,  turn  in 
the  negative  sense. 

Clockwise  is  minus  [ — ]. 
Counter-clockwise  is  plus  [-|-]. 

39.  A   bi-radial   looked   at  with    special   reference  to  the 
magnitude  and  sense-of-turning  of  a  ray's 
rotation  from  one  of  its  rays  into  the  other, 
is  called  an  angle. 

Thus,  though  we  consider  no  turning 
beyond  one  complete  rotation,  yet  the 
same    bi-radial    is    four   different   angles, 


6  ELEMENTARY  SYNTHETIC  GEOMETRY. 

T^A^  ab,  ^  —  ab,  ^  -f~  ^^y  t-  —  ba,  where  the  turning  is  always 
from  the  first-mentioned  ray  into  the  second. 

40.  If  O  (Fig.  14)  is  the  origin  or  initial  point  of  ray  a  and 
of  ray  b,  and  A  any  other  point  oh  a,  and  B  on  b,  then  ^  ■\-  ab 
may  be  written  -a^-^^  OA/OB,  or  even  -4.-^- AOB^\\\\^x&  the 
order  of  the  letters  denotes  that  the  angle  is  generated  by  a 
ray  rotating  about  O  from  OA  to  OB,  and  the  sign  fixes  the 
sense  of  that  rotation. 

/''     'x  41-  If  a  ray,  a,  is  turned  about  the  initial 

ih    c   a\ point,  C,  until  it  coincides  with  the  continu- 
ation, b,  of  its  trace  beyond  C,  the  angle  ab 
is  called  a  straight  angle. 
/"     X  42.  If  we  turn  still    more,  until  the 


c 


moving  ray  has  made  a  complete  rota- 

\^^^y  tion,   and   coincides  with   its  trace,   the 

^"^-  ^7-  angle  is  called  Si  perigon. 

43.  If  ^ab  equals  a  perigon,  then  the  ray  a  coincides  with 
the  ray  b. 

44.  When  a  bi-radial  is  looked  upon  as  an  angle,  its  two 
rays  are  called  the  arjns  of  the  angle. 

45.  Two  angles  are  eqtialM  they  can  be  so 
placed  that  their  arms  and  therefore  their  ver- 
tices coincide,  and  that  both  are  described 
simultaneously  by  the  turning  of  the  same  ray 
about  their  common  vertex. 

Fig.  18. 

46.  Equality  implies  that  both  angles  have  the  same  sense. 

47.  Two  angles  which  can  be  made  equal  by  changing  the 
sign  of  one,  are  said  to  be  equal  in  magni- 
tude but  opposite  in  sense. 

48.  Since  turning  the  plane  of  a  bi-radial 
through    half    a    revolution    changes    the 
'°'  '^"  sense  of  each  of  its  four  angles,  therefore, 

if  one  angle  by  folding  over  along  an  axis  is  made  equal  to 


PRIMARY  CONCEPTS. 


another,    then    the    angles    were    equal    in    magnitude    but 
opposite  in  sense. 

49.  Assumption    IV.    All    straight    angles   are    equal    in 
magnitude. 

50.  As  a  consequence,  all  perigons  are  equal  in  magnitude. 

51.  If  two  angles  have  a  vertex  and  an  arm 
in  common,  they  are  called  adjacent  angles. 

Fig.  20. 

52.  When  two  adjacent  angles  are  of  the  same  sense,  and 
so  situated  that  they  cannot  be  simultaneously  described,  even 


Fig.  21.  Fig.  22. 

in  part,  by  the  same  ray  rotating,  their  sum  is  the  angle  of 
like  sense  whose  arms  are  their  two  non-coincident  arms. 

53.  When  the  sum  of  any  two  angles         \^^      _^ 
is  a  straight  angle,  each  is  said  to  be  ^>\        \ 
the  supplement  of  the  other.  Fig.  23. 

54.  When  the  sum  of  any  two  angles  is 
a  perigon,  each  is  said  to  be  the  explement 
of  the  other. 

Thus  -4.-^-  ab  and  i.  -j-  ba  are  exple- 
mental.  p,^  ^^ 


CHAPTER   II. 

THE   CIRCLE. 


55.  If,  in  a  plane,  a  sect  turns  about  one  of 
its  end  points  the  other  end  point  describes  a 
curve  called  a  circle. 


56.  The  fixed  end  point  is  called  the  center  of  the  circle. 


57.  Any  sect  from  the  center  to  a  point  on 
the  curve  is  called  a  radius. 

58.  All  radii  are  equal,  being  equal  to  the 
generating  sect. 


Fig.  26. 


59.  Since  the  moving  sect,  after  rotating  through  a  perigon, 
returns  to  its  trace,  therefore  the  moving  end  point  describes 
a  closed  curve. 

60.  This  curve  divides  the  plane  into  two 
parts,  one  of  which  is  finite  and  is  swept  over 
by  the  moving  sect. 

61.  This  finite  plane  surface  is  called  the 
surface  of  the  circle.     Any  point  in  this  finite 

FioTay.  plane  is  said  to  lie  within  the  circle. 

62.  Assumed  Construction  II.  A  circle  may  be  described 
from  any  given  point  as  center  with  any  given  sect  as  radius. 

63.  A  theorem  is  a  statement  usually  capable  of  being 
inferred  from  other  statements  previously  accepted  as  true. 


THE   CIRCLE.  g 

t 

64.  A  corollary  to  a  theorem  is  a  statement  whose  truth 
follows  readily  from  that  of  the  theorem. 

65.  A  theorem  consists  of  two  parts,  the  hypothesis  (that 
which  is  assumed),  and  the  conclusion  (that  which  is  asserted 
to  follow  therefrom). 

66.  A  problem  is  a  proposition  in  which  something  is 
required  to  be  done  by  a  process  of  construction. 

67.  The  treatment  of  a  problem  in  elementary  geometry 
■consists, — 

[i]  Construction.  In  indicating  how  the  ruler  and  compasses 
are  to  be  used  in  effecting  what  is  required. 

[2]  Proof.  In  showing  that  the  construction  so  given  is 
correct. 

[3]  Determination.  In  fixing  whether  there  is  only  a  single 
solution,  or  suitable  result  of  the  indicated  construction  ;  or 
more  than  one ;  and  in  discussing  the  limitations,  which  some- 
times exist,  within  which  alone  the  solution  is  possible. 

68.  Our  assumed  constructions  allow  the  use  of  the  straight- 
edge not  marked  with  divisions,  for  drawing  and  producing 
^e^ts,  and  the  use  of  compasses  for  drawing  circles  and  the 
transference  of  sects.  It  is  important  to  note  the  implied 
restriction,  namely,  that  we  work  in  the  plane,  and  that  no 
construction  in  elementary  geometry  is  allowable  which  cannot 
be  effected  by  combinations  of  these  two  primary  construc- 
tions. 

69.  Theorem.  The  sect  to  a  point,  from  the  center  of  a 
circle,  is  less  than,  equal  to,  or  greater  than 
the  radius,  according  as  the  point  is  within, 
on,  or  without  the  circle. 

Proof.  For  any  point  Q,  within  the  cir- 
cle, lies  on  some  radius,  OQR.  If  ^  is 
without  the  circle,  then  the  sect  OS  con-  """""■fIgTIs. 

tains  the  radius  OR. 

70.  Inverse.  A  point  is  within,  on,  or  without  the  circle. 


lO 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


according  as  its  sect  from  the  center  is  less  than,  equal  to,  or 
greater  than  the  radius. 

71.  Theorem.  Circles  of  equal  radii  are  congruent. 

Proof.  For,  if  put  in  the  same  plane,  with  centers  in  coin- 
cidence, every  point  of  each  is  on  the  other,  because  of  the 
equality  of  their  radii.     Thus  Q)C\r\  ^  Q)0  [r]. 

72.  Corollary.  A  circle  turned  about  its  center  slides  on  its 
trace. 

This  fundamental  property  of  this  curve  enables  us  to  turn 
any  figure  connected  with  the  circle  about  the  center  without 
changing  the  relation  to  the  circle. 

73.  Circles  which  have  its  same  center  are  called  concentric. 

74.  Concentric  circles  with  a  point  in  common  coincide. 
75.  A  sect   whose   end  points    are    on    the 

circle  is  called  a  chord. 

"jG.  Any  chord  through  the  center  is  called 
a  diameter. 

jj.  All  diameters  are  equal,  each  being 
equal  to  two  radii. 

78.  Every  diameter  is  bisected  by  the  center 
of  the  circle. 

79.  No  circle  can  have  more  than  one 
center. 

For,  if    it   had    two,  the    diameter  through 
Fig.  30.  them  would  have  two  mid  points. 


Fig.  31. 


Fig.  32. 


80.  Any  ray  from  the  center  of  a  circle  cuts  the  circle  in 
one,  and  only  one,  point. 


THE   CIRCLE. 


II 


Fig.  33. 


81.  Any  straight  through  its  center  cuts  the  circle  in  two 
and  only  two  points. 

82.  Any  piece  of  a  circle  is  called  an  arc. 

83.  When  the  end  point  of  a  radius  de- 
scribes an  arc,  the  radius  rotates  through  an 
angle  having  its  vertex  at  the  center.  This 
angle  is  called  the  angle  at  the  center,  and  is 
said  to  be  subtended  by  the  arc  simultaneously 
described,  or  to  stand  upon  that  arc. 

84.  An   arc,    being   described    by   the   end 
point  of  a  rotating  radius,  is  said  to  have  the  same  sense  as 
the  angle  through  which  that  radius  rotates. 

85.  Arcs  congruent  and  of  the  same  sense  are  called  equal. 

86.  The  sum  of  two  arcs,  of  the  same 
circle,  or  of  equal  circles,  is  the  arc  which 
subtends  an  angle  at  the  center  equal  to  the 
sum  of  the  angles  subtended  by  those  arcs 
separately. 

87.  Theorem.  Equal  arcs  subtend  equal 
angles  at  the  center,  and,  inversely,  equal 
angles  at  the  center  stand  upon  equal  arcs. 

Proof.  For,  if  arc  AB  equal  arc  CD,  we 
may  slide  the  arc  AB,  together  with  the  radii 
OA  and  OB,  along  the  circle  until  A  coincides 
with  C\  then  will  B  coincide  with  D,  since  arc 
CD  equals  arc  AB. 

Therefore  4-  A  OB  will  coincide  with 
^  COD,  and  will  be  equal  to  it  in  magnitude  and  sense. 

88.  It  follows,  that  if  y^,  B,  C,  etc.,  denote 
points  on  the  circle  and  a,  b,  c,  etc.,  the  radii 
drawn  to  those  points,  then  every  equation 
between  arcs  AB,  BC,  etc.,  will  carry  v/ith  it  c' 
an  equation  between  the  corresponding  angles 
ab,  be,  etc. ;  and  inversely.  p^^  ,g 


Fig.  35. 


12  ELEMENTARY  SYNTHETIC  GEOMETRY. 

89.  Theorem.  In  the  same  or  equal  circles,  of  two  unequal 
arcs,  the  greater  subtends  the  greater  angle  at 
the  center. 

Proof.  If  the   first  arc   is  greater  than  the 
second,  it  equals  the  second  plus  a  third  arc, 
and   so  the   angle  which   the  first  subtends  is 
^^^^"^  greater  than  the  angle  which  the  second  sub- 

tends  by  the  angle    which     the   third    arc    subtends   at    the 
center. 

90.  Inversely :  Of  two  unequal  angles  at  the  center,  the 
greater  intercepts  the  greater  arc. 

91.  Two  arcs  which  together  equal  the  whole  circle  are 
called  explemental. 

Thus  the  explemental  angles  at  the  center  of  a  circle, 
whose  arms  are  the  same  radii,  are  said  to  stand  upon  the 
explemental  arcs  which  would  be  described  simultaneously 
Avith  the  angles,  the  greater  angle  upon  the  greater  arc. 

92.  Explemental  arcs  equal  in  magnitude  are  called  semi- 
circles. 

93.  A  semicircle  subtends  a  straight  angle.  For  two  sub- 
tend a  perigon,  and  are  equal. 

94.  Any  straight  through  the  center  cuts  the  circle  into  two 
semicircles.  For  it  makes  at  the  center  straight  angles  which 
together  are  subtended  by  the  whole  circle. 

95.  If  we  fold  over  about  a  straight  through  the  center  of 
a  circle,  the  semicircles  it  makes  are  brought  into  coinci- 
dence. 

For  every  point  on  the  turned  semicircle  must  fall  on  some 
point  of  the  other,  as  its  sect  from  the  center  is  a  radius. 

96.  Two  arcs  which  together  equal  a  semi- 
circle are  called  supplemental. 

97.  Half  a  straight  angle  is  called  a  right 
angle. 


Fic?3&.  98.  All  right  angles  are  equal  in  magnitude. 


THE   CIRCLE. 


13 


99.  The  arc  subtending  a  right  angle  is  called 
a  quadrant.     It  is  one  quarter  of  a  circle. 


100.  Two  straights  which  make  a  right 
angle  are  said  to  be  perpendicular  to  one 
another. 

10 1.  Two  angles  whose  sum  is  a  right 
angle  are  caW^d  complement al. 


Fig.  40. 


102.  An  angle  less  than  a  right  angle  is  called  acute. 

103.  An  angle  greater  than  a  right 
angle,  but  less  than  a  straight  angle,  is 
called  obtuse. 

104.  An  angle  greater  than  a  straight 
angle,  but  less  than  a  perigon,  is  called 
reflex. 

105.  An  angle  which  is  either  acute, 
right,  or  obtuse,  is  called  a  minor  angle.  Fig.  42. 

106.  An  arc  less  than  a  semicircle  is  called  a  minor  arc. 

107.  An   arc  less  than   a  circle,  but  greater  than  a  semi- 
circle, is  called  a  major  arc. 

108.  Theorem.  If  two  circles  have  one  common  point  not 
on  the  straight  through  their  centers, 
they  have  also  another  such  point. 

Proof.  Let  O  C  and  O  O  have  the 
the  point  A  in  common.  Fold  the 
figure  over  along  the  straight  through 
their  centers,  CO.  Then  the  semi- 
circles which  have  A  in  common  are 
brought  into  coincidence  with  the  other  semicircles, 
fore  these  also  have  a  common  point.  A'. 


There- 


14 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


109.  Theorem.  If  two  circles  have  a  common  point  not  on 
the   straight    through    their   centers, 
and     therefore    another   such   point, 
then  the  center-straight    bisects    the 
angles  made  at   the  centers  by  the 
radii  to  these    two    common   points, 
and  is  the  perpendicular  bisector  of 
the  common  chord. 
Proof.  For  by  folding  over  along  CO  we  bring  A  into  coin- 
cidence with  A'.     Therefore  sect  AM  =  sect  A' M.     ^  OMA 
=  t  A' MO.     t  MCA  =:  ^  A' CM.     4-  AOM  =  :^  MOA'. 


Fig.  44. 


CHAPTER    III. 


THE   FUNDAMENTAL   PROBLEMS. 


Fig. 


I  lo.  Problem.  To  bisect  any  given  sect. 

Construction.  With  its  end  points,  A  and  A',  as  centers, 
and  itself  as  radius,  describe  two 
circles.  They  will  have  one  common 
point  not  on  their  center  straight, 
and  therefore  a  second  such.  Join 
these  two  common  points,  C  and  O. 
Then  CO  bisects  the  given  sect  AA'. 

Proof.  For  A  A'    is   J_   to  CO,    and 
if   from  C  and  O  as  centers,  with   radii 
equal  to  AA' ,  two  circles  were  described,  then  A  A'  would  be 
a  common  chord,'  bisected  by  the  center-straight  CO. 

III.  Theorem.  The  straight  through  the   mid  point  of  a 
chord,    and    the    center  of   the  circle,    is    per- 
pendicular to  the  chord,  and  bisects  the  exple- 
mental  arcs,  and  their  angles  at  the  center. 

Proof.  A  and  B  are  any  points  on  qO. 
Turn  the  whole  figure  over  and  apply  it  to  its 
trace,  so  that  O  falls  on  O,  but  A  on  the  trace 
of  B,  and  B  on  the  trace  of  A.  Then  the  bisection  point,  C, 
of  the  chord  AB  falls  on  its  own  trace,  and  consequently  the 
whole  change  amounts  only  to  half  a  revolution  of  the  figure 
about  the  straight  CO. 


Fig.  46. 


15 


i6 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


l\2.  Problem.  To  bisect  any  given  angle. 

Construction.  With    its   vertex,  C,  as- 
center,  and  any  sect,  r,  as  radius,  describe 
a  circle  cutting  the  arms  of  the  angle  at 
A   and  A'.     Bisect  the  chord  AA' ,  and 
Fig.  47.  join  its  mid  point,  M,  to  the   center  C 

Then  MC  bisects  t  ACA'. 

113.  Problem.    At  a  given  point  on  a  given  straight,   to- 
draw  a  perpendicular  to  that  straight. 

Construction.  Bisect  the  straight  angle  at  the  point. 

114.  Problem.    Through   a    given    point,   not    in  a   given: 
straight,  to  draw  a  perpendicular  to  that  straight. 

Construction.  In  the  hemiplane  not  contain- 
ing the  given  point,  C,  take  any  point  D.  Call 
A  and  A'  points  where  (dC\CD~\  cuts  the  given 
straight.  Bisect  the  chord  A  A'  at  M.  Then  is 
CM ]^\.o  AA'. 

Determination.  Through  a  given  point  only 
one  perpendicular  can  be  drawn  to  a  given 
straight.  For,  if  the  plane  were  folded  over  along  the  given 
straight,  the  given  point  would  fall  on  the  production  of  any 
perpendicular  from  it  to  the  straight. 

115.  Since  the  perpendicular  from  the  center  to  a  chord  of 
a  circle  bisects  that  chord,  and  also  the 
explemental  arcs  and  the  explemental 
angles  pertaining  to  that  chord,  therefore 
the  r't  bi'  of  any  chord  passes  through  the 

A  center ;  and  the  straight  which  possesses 
any  two  of  these  seven  properties  possesses 
also  the  other  five. 

116.  Problem.    To   bisect   any  given 
arc. 

Construction.  Join  its  extremities,  and 


Fig.  49. 


\ 


X 


Fig.  50. 


draw  the  r't  bi'  of  this  chord. 


THE  FUNDAMENTAL  PROBLEMS. 


17 


117.  Theorem.  A  straight  cannot  have  more  than  two 
points  in  common  with  a  circle. 

Proof.  For,  if  it  had  a  third,  then,  since  the  r't  bi'  of  a 
chord  contains  the  center,  there  would  be  three  perpendiculars 
from  the  center  to  the  same  straight. 

118.  Theorem.  Every  point  which  joined 
to  two  points  gives  equal  sects  is  on  the  per- 
pendicular bisector  of  the  sect  joining  those 
two  points. 

Proof.  The  r't  bi'  of  the  chord  contains  the 
center. 

119.  Corollary.  Circles  with  three  points  in  common 
coincide. 


CHAPTER    IV. 


iM 


A 
Fig.  52. 


SYMMETRY   AND   SYMCENTRY. 

120.  Two  points  are  said  to  be  sym- 
metrical with  regard  to  a  given  straiglit, 
called  the  Axis  of  Symmetry,  when  the  axis 
bisects  at  right  angles  the  sect  joining  the 
tw'o  points. 


121.  Two  points  have  always  one,  and  only  one,  symmetry 


axis. 

122.  A  point  has,  with  regard  to  a  given  axis  of  symmetry, 
always  one,  and  only  one,  symmetrical  point  ;  namely,  the  one 
on  the  ray  from  the  given  point  perpendicular  to  the  axis, 
which  ends  the  sect  bisected  by  the  axis. 

123.  Two  figures  have  an  axis  of 
symmetry  when,  with  regard  to  this 
straight,  every  point  of  each  has  its 
symmetrical  point  on  the  other. 

124.  Two  figures  are  symmetrical 
when  they  can  be  placed  so  as  to  have 
an  axis  of  symmetry. 

125.  One  figure  has  an  axis  of  symmetry 
when,  with  regard  to  this  straight,  every  point  of 
the  figure  has  its  symmetrical  point  on  the  figure. 

126.  One  figure  is  symmetrical  when  it  has  an 
axis  of  symmetry. 

127.  Any  figure  has,  with  regard  to  any  given 
straight  as  axis,  always  one,  and  only  one,  sym- 
metrical figure. 

128.  One  figure  is  symmetrical  when  it  has  an  axis  with  re- 
gard to  which  its  symmetrical  figure  coincides  with  itself. 

18 


Fig.  53. 


^3\ 


Fig.  s4. 


SYMMETRY  AND   SYM GENTRY. 


19 


'C\ 


Fig.  56. 


T29,   Every  point  in  the  axis  is  symmetrical  to  itself. 

130.  The  axis  is  symmetrical  with  regard  to  itself. 

131.  Two  points  are  said   lo  be  symcentral     . 

with  regard  to  the  mid  point  of  their  joining    ~~~       '  ' 

sect.  ^"^-  55- 

132.  A  point  has,  with  regard  to  a  given  symcenter,  always 
one,  and  only  one,  symcentral  point ;  namely,  the  one  on  the 
ray  from  the  given  point  through  the  symcenter,  which  ends 
the  sect  bisected  by  the  symcenter. 

133.  Two  figures  have  a  sym- 
center  when,    with    regard    to    this 

point,  every  point  of  each  has   its   pi^_/_ .\<clC_— /^—^a 

symcentral  point  on  the  other. 

134.  Two  figures  are  symcentral 
when  they  can  be  placed  so  as  to 
have  a  symcenter. 

135.  One  figure  has  a  symcenter 
■when,  with  regard  to  this  point,  every 
point  on  the  figure  has  its  symcentral 
point  on  the  figure. 

136.  One  figure  is  symcentral  when  it 
Tias  a  symcenter. 

137.  Any  figure  has  with  regard  to  any  given  point  as  sym 
center,  always  one,  and  only  one,  symcentral  figure. 

138.  One  figure  is  symcentral  when  it  has  a 
point  with  regard  to  which  its  symcentral  figure 
coincides  with  itself. 

139.  Theorem.  A  straight,  or 
sect,  or  angle,  in  one  of  two  sym- 
metrical [or  symcentral]  figures, 
has  a  symmetrical  [or  sym- 
central] straight,  or  sect,  or  angle, 
in  the  other. 

Proof.   For  by  half   a   revolu-  Fig%9. 


v^\ 


AC 


7 


Fig.  57. 


D' 


J 


^^^fi^ 


Fig.  58. 


20 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


tion   [rotation]  of  one  figure  about  the  axis  [symcenter]  the 

two  are  made  to  coincide. 

140.  If  point  C  •!•  C ,  and  from  C 
rays  a  and  b  •!•  rays  a  and  b'  from  C , 
the  -4-  -\-  ab  '\'  4-  —  (^'b' ,  remembering- 
that  in  any  angle  the  turning  is  from  the 
first-mentioned  ray  into  the  second,  and 

the  sign  denotes  the  sense  of  that  turning. 

141.  The  intersection  point  of  two  straights  is  symmetrical 

[or  symcentral]  to  the  intersection  of  two  symmetrical  [or  sym- 

central]  to  those. 


Fig.  60. 


Fig.  61.  Fig.  62. 

142.  The  intersection  point  of  two  symmetrical  straights  is 
on  the  axis. 

143.  If  three  points  lie  in  a  straight,  their  symmetrical  [or 
symcentral]  points  lie  in  a  symmetrical  [or  symcentral] 
straight. 


Fig.  64. 


SYMMETRY  AND   SYMCENTRY. 


21 


Fig.  66. 


T4.I.  The  bisector  of  an  angle  is  symmetrical  [or  sym- 
central]  to  the  bisector  of  the  symmetrical  [or  symcentral]  angle. 

145.  Every  point  symmetrical  [or  symcentral]  to  itself 
lies  in  the  axis  [or  symcenter], 

146.  The  angle  between  two  sym- 
metrical straights  is  bisected  by  the  axis. 

147.  Any  straight  is  symmetrical    with  regard  to    I 
any  of  its  perpendiculars.  r 

148.  Any   straight    is  symcentral   with   regard   to 
any  of  its  points. 

Thus  the  intersection  point  of  two 
straights  is  a  symcenter  for  each ;  so  the 
non-adjacent  or  vertical  angles  are  equal, 
and   their  bisectors,   being  symcentral  rays  Fig.  67. 

from  the  symcenter,  are  in  one  straight. 

149.  Theorem.  Two  intersecting  straights  are  symmetrical 
with  regard  to  either  of  their  angle  bisectors. 

Proof.  For  the  points  which  would  be  brought  into  coin- 
cidence by  folding  along  this  bisector  were  symm.etrical  with 
regard  to  it. 

1 50.  Any  circle  is  symmetrical  with*  regard  to  any  of  its 
diameters,  and  symcentral  with  regard  to  its  center. 


Fig.  69.  Fig.  70. 

151.  Every  axis  of  symmetry  of  a  circle  passes  through  the 
center. 

For  the  diameter  perpendicular  to  this  axis  is  bisected  by  it. 

152.  A  figure  made  up  of  a  straight  and  a  point  without  it 
is  symmetrical,  but  never  symcentral. 


CHAPTER   V. 


TANGENTS. 

153.  Theorem.  Every  point  on  the  per- 
pendicular bisector  of  a  sect  is  the  center  of  a 
circle  passing  through  its  end  points. 
For  A  -l-A',  axis  d;  .\  CA  =  CA\ 
Thus  sects  from  any  point  on  its  perpendi- 
cular bisector  to  the  end  points  of  the  sect  are 
equal. 

154.  A  straight  which  has  two  points  in  common  with  a 
circle  is  called  a  secant. 

155.  A  straight  which  has  only  one  point  in  common 
with  a  circle  is  called  a  tangent  to  the  circle,  and  the  point 
is  called  the  point  of  contact. 


Fig.  71. 


156.  Since  any  chord  is  bisected  by  the 
perpendicular  from  the  center,  .*.  a  straight 
X  to  a  diameter  at  an  end  point  has  only  this 
point  in  common  with  the  circle. 

This  point  of  the  circle  is  symmetrical  to 
itself  with  regard  to  this  diameter  as  axis. 
But  if  we  draw  through  this    point  C  any 


TANGENTS. 


23 


straight  j^,  not  J_  to  the  radius,  then  the  perpendicular  from  the 
center,  O,  will  meet  this  straight  s  at  some  other  point,  F- 
Hence  the  straight  s  cuts  the  circle  again  at  C  •\'C,  axis  OF- 
Therefore : 

Theorem.  At  every  point  on  the  circle  one,  and  only  one, 
tangent  can  be  drawn,  namely,  the  perpendicular  to  the  radius 
at  the  point. 

157.  The  perpendicular  to  a  tangent  from  the  center  of  a 
circle  cuts  it  in  the  point  of  contact. 

158.  The  perpendicular  to  the"  tangent  at  the  point  of  con- 
tact contains  the  center. 

159.  The  radius  to  the  point  of  contact  of  a  tangent  is  per- 
pendicular to  the  tangent. 

160.  To  draw  the  tangent  to  a  circle  at  any  point,  draw  the 
perpendicular  to  the  radius  at  that  point. 

161.  Let  6>  be  a  point  not  in  the  straight  s,  and  OC X.  to  s  : 
then  s  is  tangent  to  O^  \PC^  at  C. 

Any  second  circle  concentric  with  the 
first,  but  of  lesser  radius,  lies  wholly  within 
the  first. 

A  third  concentric  circle,  with  radius 
>  C'^T,  lies  wholly  without  the  OO  \PC\ 
and   cuts  s  in  D-\-D',  axis   OC;    .'.  CD  =  fig.  75. 

CD'. 

A  fourth  concentric  circle,  with  radius  >  OD,  lies  wholly 
without  the  third ;  .*.  its  intersections  with  s  lie  without  the 
sect  DD'. 

Hence  the  four  following  theorems  : 

162.  A  straight  will  be  a  secant,  a  tangent, 
or  not  meet  the  circle,  according  as  the  perpen- 
dicular to  it  from  the  center  is  less  than,  equal 
to,  or  greater  than  the  radius. 

163.  The  perpendicular  is  the  least  sect 
between  a  given  point  and  a  given  straight.  f,^.  yg. 


24 


ELEMENTARY  SYNTHETIC  GEOMETRY 


Fig.  77. 


164.  Except  the  perpendicular,  any  sect  from  a  poirit  to  a 
straight  is  called  an  obligtie. 

165.  Two  obliques  from  a  point  to  a  straight,  making  equal 
sects  from  the  foot  of  the  perpendicular,  are  equal. 

166.  Of  any  two  obliques  between  a  point  and  a  straight, 
that  which  makes  the  greater  sect  from  the  foot  of  the  perpen- 
dicular is  the  greater. 

167.  Problem.  From   a  given  point  without   the  circle  to 
dravv^  a  tangent  to  the  circle. 

Construction,  Join  the  given  point  A  to 
the  center  C,  cutting  the  circle  in  B.  Draw 
BDX_  to  CB,  and  cutting  in  D  the  Q)C[CA]. 
Join  DC,  cutting  qC  {CB']  in  F.  Then  AF  \s 
tangent  \.o  Q)C  ICB\ 

Proof.  Radius  CA,  _L  to  chord  HD,  bisects 
arc  HD\  .'.  if  we  rotate  the  figure  until  H  comes  upon  the 
trace  of  A,  then  A  is  on  the  trace  of  D,  .*.  tangent  HB  on  trace 
oi  AF. 

Determination.  Always  two  and  only  two  tangents. 

168.  Corollary.   By  symmetry  the  straight  through  C,  the 

center  of  a  circle,  and  A  an  exter- 
nal point,  bisects  the  angle  be- 
tween the  two  tangents  from  A  to 
OC,  and  also  the  angle  between 
the  radii  to  the  points  of  contact, 
F  and  F' ;  and  sect  AF  =  sect 
AF'. 

169.  Corollary.  Any  point  from 
which  the  perpendiculars  on  two  intersecting  straights  are 
equal,  is  on  one  of  their  angle  bisectors.  It  is  center  of  a 
circle  to  which  they  .are  tangent. 

170.  Corollary.  The  centers  of  all  circles  tangent  to  two 
intersecting  straights  are  in  their  angle  bisectors. 

171.  Inversely.  From  any  point  on  a  bisector  of  an  angle 


Fig.  78. 


TANGENTS.  2$ 

made  by  two  straights,  the  perpendiculars  to  those  straights 
are  equal. 

For  the  bisector  is  a  symmetry  axis  for  the  two  straights  : 
so  when  we  fold  along  it,  the  foot  of  the  perpendicular  to  one 
straight  falls  on  the  other  straight,  and  there  is  only  one  per- 
pendicular from  a  point  to  a  straight. 


CHAPTER   VI. 

CHORDS. 

172.  Take  AB  any  chord  in  oO.  The  qA  [AB]  cuts  OO- 
in  two  points,  B \ B' ,  axis  AO  [the  center- 
straight].  But  the  end  points  of  all  sects  from 
A  which    are    equal   to   AB   must  lie    on  Q)A 

^°  {_AB],  ;   whence  : 

Theorem.  Chords  from  any  point  of  a  circle 
are  equal   in  pairs,  one    on   each    side    of    the 
Fig' 79/        diameter  from  that  point. 

173.  A  circle  =  ©(9,  and  containing  a  chord  =  AB,  can  be 
superimposed  upon  OO,  and  then  rotated  until  one  end  of  the 
chord  comes  at  A.  The  other  end  of  this  chord  then  lies  on 
both  QO  and  QA  [AB],  and  so  falls  on  B  ox B'  \  and  the  chord 
coincides  with  AB  or  AB' .     Hence  the  theorem: 

In  the  same  or  equal  circles,  to  equal  chords  pertain  equal 
minor  arcs. 

174.  Corollary.  In  the  same  or  equal  circles  of  arcs  pertain- 
ing to  equal  chords  any  two  are  either  equal  or  explemental. 

175.  If  with  center  A  and  radius  AC  <.  AB  we  describe  a 
second  circle,  it  will  lie  wholly  within  qA  [AB].  Conse- 
quently it  cuts  QO  in  points  C  and  C  on  the  arc  BAB' \ 
.'.  arc  AC  <  arc  AB.  Thus  if  the  chord  decreases,  so  does  the 
minor  arc ;  and  inversely,  of  two  unequal  minor  arcs,  the  greater 
has  the  greater  chord. 

176.  If  the  chord  increases,  its  major  arc  decreases,  since  its 
major  and  minor  arcs  are  always  explemental.  Inversely,  if  a 
major  arc  decreases,  its  chord  increases. 

177.  A  diameter  is  the  greatest  chord.  Every  other  chord 
equals  a  piece  of  the  diameter. 

26 


CHORDS. 


27 


178.  Equal  arcs,  being  congruent,  have  equal  chords. 
Therefore,  also,  explemental  arcs  have  equal  chords. 
179.    Equal  chords,  having  equal  minor  arcs,  which  may  be 
brought  into  coincidence  by  rotation  about  the  center,  have' 
also  equal  perpendiculars  from  the  center. 

180.  In  the  same  or  equal  circles,  chords  which  have  equal 
perpendiculars  from  the  center,  since  by  rotation  one  may  be 
put  upon  the  other,  are  equal.  S >.b 

181.  Since  the  end  point  6^  of  a  chord 
AC  <  AB  lies  on  the  minor  arc  AB,  :.  it  is 
on  the  side  of  AB  remote  from  the  center 
0\  .'.  the  mid  point  of  chord  AC  is  on  the 
side  oi  AB  remote  from  O.  fig.  80. 

Theorem.  In  the  same  or  equal  circles,  the  greater  chord 
has  the  lesser  perpendicular  from  the  center. 

182.  Inversely.  The  chord  with  the  greater  perpendicular 
from  the  center  is  the  lesser;  for  [181]  it  cannot  be  the  greater 
chord,  nor  [179]  can  they  be  equal. 

183.  Problem.  At  a  given  point  G,  in  a  given  straight  s,  to 

make  an  angle  equal  to  a  given 
^v\g\e  ACB. 

Construction.  With  any  radius 
draw  OC[r],  cutting  CA  at  Z?, 
and  CB  at  F.  Join  DF.  Draw 
OC  [r],  cutting  the  given  straight 
at  H.  Draw  QH  \r'  =  BF],  cutting  QG  [r]  at  K.  Join  GK. 
Then  ^B'GK=  ^  ACB. 

Proof.   06^  [r]  =  OA  [r],  and  chord  //'TT^  chord  DF. 
.•.  minor  arc  HK  =  minor  arc  BF. 
.'.  minor  ^  HGK^=.  minor  4-  BCF. 

Determination.  The  construction  will  give  four  minor  angles, 
at  G,  all  equal,  namely, 


Fig.  81. 


H,GK,;  K,GH,;   KJGH,-,  HfiK,. 


CHAPTER    VII. 

TWO   CIRCLES. 

184.  A  figure  formed  by  two  circles  is  symmetrical  with 
regard  to  their  center-straight  as 
axis. 

Every    chord    perpendicular    to 
this  axis  is  bisected  by  it. 

If   the    circles   have    a   common 

point  on  this  straight,  they  cannot 

^"^-  ^^-  have  any  other  point  in  common,  for 

any  point  in  each  has  its  symmetrical  point  with  regard  to  this 

axis,  and  circles  with  three  points  in  common  coincide. 

185.  Two  circles  with  only  one  point  in  common  are  called 
tangent,  are  said  to  touch  ;  and  the  common  point  is  called  the 

point  of  tangency  or  contact. 

186.  If  two  circles  touch,  then, 
since  there  is  only  one  common 
point,  this  point  of  contact  lies 
on  the  center-straight,  and  a  per- 
pendicular to  the  center-straight 
^"^'  ^^'  through  the  point  of  contact  is  a 

common  tangent  to  the  two  circles. 


CHAPTER   VIII. 

PARALLELS. 

187.  A  straight  cutting  across  other  straights  is  called  a. 
transversal. 

[In  plane  geometry,  all  are  in  one  plane.] 

188.  If,  in  a  plane,  two  straights  are  cut 
in  two  distinct  points  by  a  transversal,  at 
each  of    these    points  four  positive   minor  lo.z^. 
angles  are  made. 

Of  these  eight  angles,  four  are  between  the  two  straights^ 
[namely,  3,  4,  a,  ^],  and  are  called  Interior         \ 
Angles  :  the  other  four  lie  outside  the  two  J^ 

straights,  and  are  called  Exterior  Angles.  \  ^ 

Angles,  one  at  each   point,  which    lie  <^  d 

on  the  same  side  of  the  transversal,  the 

one  exterior  and  the  other  interior,  are  called  Corresponding 
Angles  [e.g.,  1  and  a']. 

Two  non  adjacent  angles  on  opposite  sides  of  the  trans- 
versal, and  both  interior  or  both  exterior,  are  called  Alternate 
Angles  [e.g.,  3  and  «]. 

Two  angles  on  the  same  side  of  the  transversal,  and  both 
interior  or  both  exterior,  are  called  Conjugate  Angles  [e.g.,  4 
and  a\ 

189.  Theorem.  If  two  corresponding  or  two  alternate  angles 
are  equal,  or  two  conjugate  angles  are  supplemental,  then  every 
angle  is  equal  to  its  corresponding  and  to  its  alternate,  and 
supplemental  to  its  conjugate. 

[Use  vertical  angles  and  supplemental  adjacent  angles.} 

29 


30 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


Fig.  86. 


190.  Parallels  are  straights  in  the  same 
plane  which  nowhere  meet. 

[Note.  As  we  are  working  on  a  plane, 
the  clause  "  in  the  same  plane"  would  be 
understood  even  if  not  mentioned.] 

191.  Assumption  V.  Two  coplanar 
straights  are  parallel  if  a  transversal  makes 
equal  alternate  angles. 

192.  Assumption  VU  If  two  coplanar 
straights  cut  by  a  transversal  have  a  pair 
of  alternate  interior  angles  unequal,  they 
meet  on  that  side  of  the  transversal  where 
lies  the  smaller  angle. 

193.  Theorem.  If  two  straights  cut  by 
a  transversal  have  corresponding  angles  equal,  or  conjugate 
angles  supplemental,  they  are  parallel. 

For  either  hypothesis  makes  the  alternate  angles  equal. 
194.  If  two  straights  cut  by  a  transversal  have  conjugate 
angles  not  supplemental,  they  meet. 
For  the  alternate  angles  are  unequal. 


:^ 


195.  Problem.  Through  a  given 
point  to  draw  a  parallel  to  a  given 
straight. 

Construction.  Join  the  given  point 
P  to  any  point  C,  of  the  given  straight 
CB.  Then  at  P  make  an  angle  CPD  alternate  and  equal  to 
tPCB. 

Determination.  There  is  only  one  solution. 

196.  Corollary.  Two  coplanar  straights  parallel  to  the  same 
straight  are  parallel  to  one  another. 

For  they  cannot  meet. 

197.  Theorem.    If   a   transversal   cuts   two   parallels,   the 
alternate  angles  are  equal. 


PARALLELS. 


31 


Fig.  91. 


Proof.  For  if  they  were  unequal,  the  straights  would  meet. 

198.  Theorem.    Any   two    parallels  c       x^ 

are  symcentral  with  regard  to  the  mid  \3fl 

point  of  the  sect  which  they  intercept  ^ ° 

on  any  transversal.  fig.  90. 

Proof.  Rotating  the  figure  about  M  through  a  straight 
angle  brings  A  into  coincidence  with  the  trace  of  B  and 
:4.  CAM  into  coincidence  with  the  trace  of  the  equal  alternate 
t  DBM. 

199.  Two  angles  with  their 
arms  parallel  are  either  equal  or 
supplemental  [189  and  197.] 

200.  If  two  angles  have  their  arms  respectively  perpendic- 
ular, they  are   either  equal  or  supple- 
mental. 

For  rotating  one  of  the  angles 
through  a  r't  4-  around  its  vertex,  its 
arms  become  _L  to  their  traces,  and 
.'.  II  to  the  arms  of  the  other  %.. 

201.  Points  all  in  the  same  straight 
are  called  costraight. 

202.  Problem.  To  pass  a  circle  through  any  three  points 
not  costraight. 

Construction.  Join  the  three  points  by  three  sects ;  to 
these  sects  erect  r't  bisectors ;  of  these  every 
two  will  meet,  since  they  make  an  angle  =  or 
,  supplemental  to  that  to  whose  arms  they  are 
J_.  Suppose  two  to  meet  at  C.  This  point 
joined  to  the  three  points  gives  three  equal 
sects. 

Therefore  it  is  the  center  of  a  circle  containing  the  three 
-given  points. 

203.  Corollary.  The  center  of  any  O  through  the  three 
points  must  lie  on  all  three  r't  bi's. 


Fig.  92. 


Fig.  93. 


32 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


.'.  the  third  r't  bi'  passes  through  O. 

204.  Problem.  To  describe  a  circle  touching  three  givei* 
intersecting  straights  not  all  through  the  same  point. 

Construction.  At  each  of  two  intersection  points  draw  the 
two  angle-bisectors.  Every  pair  of  these  meet,  since  they 
make  conjugate  angles  which  are  not  supplemental.     [Two  of 


the  four  different  angles  bisected  are  together  less  than  a 
straight  angle ;  the  other  two  each  less  than  a  straight  angle, 
and  the  angle  between  bisectors  of  supplemental  adjacent 
angles  is  right.] 

From  any  point,  as  /,  on  a  bisector  through  A  and  one 
through  B,  drop  a  perpendicular  upon  one  of  the  given 
straights,  as  AB.  A  circle  described  with  this  perpendicular 
as  radius  is  tangent  to  AB\  but  it  also  touches  the  second 
given  straight  BC  [/  lies  on  the  bisector  of  an  angle  between 
AB  and  ^C],  and  the  third  CA  [/  is  on  a  bisector  of  an  angle 
between  AB  and  AC]. 

Determination.  Every  intersection  point  of  two  angle- 
bisectors  has  thus  equal  perpendiculars  to  the  three  given 
straights.     It  is  therefore  on  a  third  angle-bisector. 

Thus  the  four  intersection  points  of  the  two  bisectors 
through  A  with  the  two  through  B  are  the  eight  intersection 
points  of  the  two  bisectors  through  C  with  the  other  four.. 


PARALLELS. 


33 


Thus  the  two  bisectors  through  the  third  point  give  no  new 
intersections,  and  there  are  just  four  solutions. 

205.  Problem.  To  draw  a  common  tangent  to  two  given 
circles. 

Construction.  A  and  B  are 
the  points  where  QC  [CA]  and 
OOIOBI  are  cut  by  CO.  Sup- 
pose CA  >  OB.  From  AC  or 
AO  cut  ofi  AD  =  OB.  Describe 
OC  [CD^.  To  it,  from  6?,  draw 
tangent  OP.  Let  CP  cut  qC  [CA]  in  Q.  Through  O,  on  the 
same  side  of  OP  as  Q,  draw  OR  \\  to  CP,  cutting  qO  [OB]  in 
R.     Then  QR  is  a  common  tangent. 


Fig.  95. 


Fig.  96. 

Proof.    Radii    CQ  =  CA,    CP=CD',     .-.    PQ  =  AD - 

OB  =  OR. 

But  OR  I  to  PQ  and  ^  OPQ  a  r't  ^  ;  .-.  ^  /'C>y?  is  a 
r't  ^. 

.-.  e  -I-  y?,  axis  _L  to  OP,  .'.  0P\\  to  QR,  .'.  t  PQR  = 
4.  QRO  =  a  r't  ^. 


CHAPTER   IX. 


THE   TRIANGLE. 

206.  Three  points  A,  B,  C,  not  co-straight, 
and  the  three  straights  they  determine,  form 
a  figure  called  a  triangle. 

207.  The  three  points  of  intersection  are 
the  three  vertices   of  the  triangle  \_A,  B,  C, 

208.  The  circle  through  the  vertices  of  a 
triangle  is  called  its  circumcircle,  O  O  [R\,  and 
the  center  O  of  the  circumcircle  is  called  the 
circumcenter  of  the  triangle ;  its  radius,  R,  the 
circumradius. 

209.  The  three  sects  joining  the  vertices  are 
the  sides  of  the  triangle.  The  side  opposite  the  angle  A  is 
called  a ;  opposite  :4-  B  \s>  side  b ;  opposite  C,  c. 

210.  Straights  which  all  intersect  in  the  same  point  are 
called  concurrent. 

21  r.  The  three  perpendicular  bisectors  of  the  sides  of  a 
triangle  are  concurrent  in  its  circumcenter. 


Fig.  98. 


Fig.  99. 


THE    TRIANGLE. 


35 


212.  The  circle  tangent  to  the  three  sides  of  a  triangle  is 
called  its  in-circle,  ©/  [r],  and  its  center  /,  the  triangle's  in- 
center  [r,  in-radius]. 

213.  The  three  internal  bisectors  of  the  angles  of  a  triangle 
are  concurrent  in  its  in-center. 

214.  A  circle  touching  one  side  of  a  triangle  and  the  other 
two  sides  produced  is  called  an  escribed  cxrcle,  or  ex-Q. 

The  three  centers  /, ,  /, ,  /,  of  the  escribed  circles  O  /, 
r^i]»  O  ^1  [^j]'  O  h  [''3]  of  ^  triangle  are  called  its  ex-centers. 

215.  The  sum  of  two  sects  is  the  sect  obtained  by  placing 
them  on  the  same  straight,  with  one  end  point  of  each  in  coin- 
cidence, but  no  other  point  in  common. 

216.  An  exterior  aiigle  of  a  triangle  is  one  between  a  side  and 
the  continuation  of  another  side. 

217.  Through  the  vertex  B  of  a.  A 
draw  BD  \\  to  AC.       The   exterior  ^ 
ABE    is    made    up   of   ^  ABD  =  4.  c- 
BAC  [alternate],   and    ^  DBE  =  4. 
^C5  [corresponding].     Therefore: 

Theorem.  In  every  triangle  any  exterior  angle  equals  the 
sum  of  the  two  interior  angles  not  adjacent  to  it.     Therefore : 

218.  Theorem.  The  sum  of  the  angles  in  any  plane  triangle 
is  a  straight  angle. 

219.  Corollary.  In  a  triangle,  at  least  two  angles  are  acute. 
The  third  angle  may  be  acute,  right,  or  obtuse ;  and  the  tri- 
angle is  called  acute-angled,  right-angled,  or  obtuse-angled,  ac- 
cordingly. 

220.  In  a  right-angled  triangle  the  side  opposite  the  right 
angle  is  called  the  hypothenuse. 

221.  A  triangle  with  two  sides  equal  is  called 
isosceles. 

222.  Theorem.  If  one  side  of  a  triangle  be 
greater  than  a  second,  the  angle  opposite  the  first 
must  be  greater  than  the  angle  opposite  the  second. 


Fig.  ioi. 


Fig.  102. 


36  ELEMENTARY  SYNTHETIC  GEOMETRY. 

Given  BA  >  BC.  Draw  bisector  BD  of  ^  B,  and' 
fold  over  along  this  axis.  Then  C  falls 
on  BA  at  C  between  B  and  A.  Then 
4(;  (7  now  appears  as  an  exterior  2^  to  A 
AC'D,  and  .".  >  4-  -^  ^ot  adjacent. 
If  one  angle  of  a  triangle  is  greater  than  a 
second,  the  side  opposite  the  first  must  be  greater  than  the 
side  opposite  the  second. 

Proof.  Given  i^  C  >  4.  A.  Draw  the  bisector  BD  of 
^  B.  Then  is  :^  ADB  [=  ^  {C -\- ^B)\  >  ^  BDC  [=  ^ 
{A  -\-  ^B)] ;  therefore  on  folding  over  along  the  axis  BD,  ^ 
BDC  will  fall  within  ^  ADB,  and  therefore  C  must  fall  be- 
tween A  and  B. 

224.  Corollary  I.  In  an  isosceles  triangle,  the  angles  op- 
posite the  equal  sides  are  equal. 

225.  Corollary  II.  If  two  angles  of  a  triangle  are  equal,  the 
triangle  is  isosceles. 

226.  If  we  join  CC   in    the  preceding 
figure    then     ^    DCC  =  ^    CCD,    since 
A   DCC    is    isosceles;    .'.    2j^    ACC    <  4C 
F?G.  .04.  "  CCA  ;  .-.  AC  <  AC.     But  AC  =  AB  ~ 

BC.     Therefore : 

Theorem.  In  every  triangle  the  difference  of  two  sides  is 
less  than  the  third  side. 

AB  -BC  <AC;  .'.  AB  <  AC+BC.     Therefore: 
227.  Theorem.   In  every  triangle  the  sum  of  any  two  sides 
is  greater  than  the  third  side. 


CHAPTER   X. 

POLYGONS. 

228.  A  number  of  sects,  the   second  beginning  at  the  end 
point  of  the  first,  the  third  at  the  end  point  of 
the  second,  etc.,  are  called  a  broken  line, 

229.  Theorem.  The  sect  between  two  given 
points  is  smaller  than  any  broken  line  between       ^'°*  "^" 
the  points. 

Proof.  Beginning  at  one  of  the 
points,  reduce  the  number  of  sects  in  the 
broken  line,  and  its  size,  by  substituting 
for  the  first  two  the  sect  joining  their 
non-coincident  end-points.      So  proceed  ^^'  '   ' 

until  the  sect  between  the  two  given  points  is  attained. 

230.  If,  in  a  broken  line,  the  beginning  point  of  the  first 
sect  coincides  with  the  ending  point  of  the  last,  the  figure  is 
called  2.  polygon,  the  broken  hne  its  perimeter,  and  the  sects  its 
sides. 

231.  A  polygon  has  as  many  angles  as  it  has  sides. 

232.  A  polygon,  no  side  of  which  cuts  another,  is  called  an 
undivided  polygon. 

233.  In  a  plane,  the  perimeter  of  ain  undivided  polygon  en- 
closes one  finite  uncut  piece,  which  is  called  the  surface  of  the 
polygon. 

234.  By  the  angles  of  an  undivided  polygon  we  under- 
stand those  each  described  by  a  ray  sweeping  over  part  of  the 
surface  of  the  polygon. 

37 


38 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


235.  An  undivided  polygon   each  of  whose  angles  is  less 
than  a  straight  angle  is  called  convex. 

D  R  o 


Fig.  108. 


Fig.  109. 


236.  Any  sect  joining  vertices  not  consecutive  is  called  a 
diagonal  of  the  polygon. 

237.  A  polygon  of  three  angles  is  a  trigon  or  triangle  ;  one 
of  four  angles  is  a  tetragon  ;  of  five,  a  pentagon ;  of  six,  a 
hexagon  ;  of  seven,  a  heptagon  ;  of  eight,  an  octagon  ;  of  nine, 
a  nonagon ;  of  ten,  a  decagon  ;  of  twelve,  a  dodecagon ;  of 
fifteen,  a  quindecagon. 

238.  By  the  word  quadrilateral,  unqualified,  we  will  mean  an 
undivided  tetragon. 

239.  A  polygon  both  equilateral  and  equiangular  is  called 
regular. 


Fig. 


Fig.  III. 


240.  A  regular  polygon  whose  sides  intersect  is  called  a 
star  polygon. 

241.  A  regular  polygon,  if  undivided,  is  convex. 
242.  Theorem.  In  a  plane,  the  sum  of  the 

angles  of  an   undivided   polygon  is  two  less 
straight  angles  than  it  has  sides. 
Fjg.  112.  Proof.  By  a  diagonal  within  the  polygon 

cut  off  a  triangle.    This  diminishes  the  number  of  sides  by  one. 


POL  YGONS. 


39 


and  the  sum  of  the  angles  by  a  straight  angle.  So  reduce  the 
sides  to  three.  We  have  left  two  more  sides  than  straight 
angles. 

243.   If  through  its  second  end  point  we  produce  every  side 
of  a  convex  polygon,  we  get  an  exterior  angle  at 
every  vertex.     This  angle  is  the  supplement  of 
the  adjacent  angle  in  the  polygon ;  therefore  : 

Theorem.  In  any  convex  plane  polygon  the 
«5am  of  the  exterior  angles,  one  at  each  vertex,  is 
a  perigon. 

243  (b).  A  trapezoid  is  a  quadrilateral  with  two  sides  parallel. 


CHAPTER    XI. 


PERIPHERY   ANGLES. 


244.  An  inscribed  angle  is  one  whose  arms 
are  chords  from  the  same  point  on  the  circle. 

245.  A  tanchord  angle  is  one  between  a 
tangent  to  a  circle  and  a  chord  from  the  point 
of  contact. 

246.  Inscribed  angles  and  tanchord  angles 
are  CdiWed  periphery  angles. 

247.  A  periphery  angle  is  said  to  intercept  or 
stand  upon  the  arc  swept  over  by  the  describing 
ray. 

248.  Theorem.  A  periphery  angle  is  half  the 
angle  at  the  center,  standing  on  the  same  arc. 

Proof.  Draw  the  bisector  of  the  minor  [reflex] 
angle  at  the  center  on  the  minor  [major]  arc  in- 
tercepted by  an  acute  [obtuse]  tanchord  angle. 
Fig.  116.  This  is  _]_  to  the  chord  ;  .•.  it  makes  with  the 
radius  to  the  point  of  contact  an  angle  whose  arms  are  A.  to 
those  of  the  tanchord  angle ;  .'.  both  being  acute  [obtuse} 
they  are  equal. 

An  inscribed  angle  is  the  difference  of  two  tanchord  angles, 
and  its  intercepted  arc  is  the  difference  of  theirs  ;  so  its  angle 
at  the  center  is  the  difference  of  theirs. 

249.  Corollary  I.  All  periphery  angles  on  the  same  arc  are 
equal. 

For  each  is  equal  to  half  the  angle  at  the  center  on  this  arc. 

40 


PERIPHERY  ANGLES. 


41 


250.  Corollary    II.  Periphery  angles  on  explemental  arcs 
are  supplemental. 

T         A  .,  Ao  A3 


Fig.  T17.  Fig.  118. 

For  they  are  halves  of  the  explemental  angles  at  the  center. 

251.    Points  on  the  same  circle  are  called  concyclic. 

2^2.  A  polygon  whose  vertices  are  concyclic  is  called  cyclic. 

253.  The  opposite  angles  are  supplemental  in  every  cyclic 
quadrilateral  (250).       • 

254.  In  a  cyclic  quadrilateral  any  angle  equals  the  opposite 
exterior  angle. 

255.  In  the  same  or  equal  circles  all  equal  periphery  angles 
intercept  equal  arcs ;  and  inversely. 

For  the  corresponding  angles  at  the  center  are  equal. 

257.  Theorem.  An  angle  made  by  two  chords  is  half  the 
-sum  of  the  angles  at  the  center  standing  on  the  arcs  intercepted 
by  it  and  its  vertical. 

Proof,     ^.x^-^-y-^-^z. 


Fig. 


Fig. 


258.  Theorem.  An  angle  made  by  two  secants  is  half  the 
difference  of  angles  at  the  center  standing  on  the  intercepted 
arcs. 

Proof,     ^x  =i  ^.y  —  ^2. 


42 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


259.  Theorem.  All  angles  equal  to  a  given  angle,  and  whose 
arms  pass  through  two  given  points,  have  their 
vertices  on  two  symmetrical  arcs  ending  in  these 
points. 

Proof.  Find  the  center  of  a  circle  through  the 
vertex  of  one  such  angle  and  the  two  given  points, 
and  draw  the  arc  ending  in  those  points  and  con- 
taining the  vertex,  and  the  arc  -I-  to  this  with  regard 
to  the  straight  through  the  given  points. 
All  angles  with  arms  through  the  two  given  points  are  < 
the  given  angle  if  their  vertices   fall  without  this  figure  [258]  ; 
>  if  within  it  [257]. 

260.  Corollary  I.  The  vertices  of  all  right-angled  triangles 
on  the  same  hypothenuse  are  concyclic. 

261.  Cor.  II.  If  two  opposite  angles  of  a  quadrilateral  are 
supplemental,  it  is  cyclic. 


Fig.  122. 


CHAPTER   XII. 

THE   SYMMETRICAL  TRIANGLE. 

262.  The  figure  consisting  of  three  points  can  only  be  sym- 
central  if  they  are  in  the  same  straight :  consequently  no  tri- 
angle has  a  symcenter. 

263.  In  any  triangle  a  sect  joining  a  vertex  to  the  mid 
point  of  the  opposite  side  is  called  a  median. 


Fig.  123. 


Fig.  124. 


264.  A  perpendicular  from  a  vertex  to  the  opposite  side  is 
called  an  altitude. 

265.  The  figure  consisting  of  three  points  can  only  be  sym- 
metrical with  regard  to  an  axis  passing  through  one  and  bisect- 
ing at  right  angles  the  sect  joining  the  other  two  ;  consequently, 
every  symmetrical  triangle  is  isosceles,  and  has  a  median  which 
is  an  altitude-  and  an  angle-bisector. 

266.  If  with  the  intersection  of  the  equal 
sides  of  any  isosceles  triangle  as  center,  and  one 
of  the  sides  as  radius,  we  describe  a  circle,  it 
will  pass  through  the  other  two  vertices. 

Therefore   in    every    isosceles   triangle    the  Fig.  125. 

median  concurrent  with  the  equal  sides  is  an  altitude-  and  an 
angle-bisector.     So  every  isosceles  triangle  is  symmetrical. 

43 


44 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


267.    Theorem.  A  triangle    having   a   median 
which  is  an  angle-bisector  is  isosceles. 

Proof.  Produce  this  median  BD  to  F,  making 
DF=BD.     ]o\n  AF.     A^/^/^  is  symcentral  to 
A   aDBC;   .:  ^F=  4.CBD,  and  FA  =  BC.      But 
^CBD  =  ^DBA  ; 

.-.  :^F=  ^DBA  ;      .-.  FA  =  AB ;      .-.  AB  =  BC. 


F'G.  126.  268(«).  Theorem.  A  triangle  is  symmetrical  if 

two  angle-bisectors  are  equal. 

p^  Proof.  If  ^OBC  is  not  =  ^OCB,  suppose 

^OBC>  4-OCB  ; 

.:  CD>  BE.     (304.) 
^CBF=^ECB; 
tBCF=  tEBC\ 
.:  BF=  CE, 
CF=BE. 
F.G.,.6W.  ]o\n  DF. 

Then,  since  BF  ■=  BD, 

.'.  ^BFD  =  ^BDF; 

^OCD  <  4.OBE, 

^COD  =  4-BOE', 

.'.  to  DO  ^OEB, 

.'.  to  DC  >  t-BFC. 


and  by  hypoth., 
and 

and 


THE   SYMMETRICAL    TRIANGLE.  45; 

Hence,  subtracting    ^BDF  —  4-BFD, 

.'.4-FDC>  tDFC\ 

,:CF>CD, 

.'.  BE  >  CD, 

.-.  BE  >  and  <  CD; 
which  is  absurd. 

.'.  :^OBC  =  ^OCB. 

268  {b).  If  any  triangle  has  one  of  the  following  properties, 
it  has  all : 

[i]  Symmetry. 

[2]  Two  equal  sides. 

[3]  Two  equal  angles. 

[4]  A  median  which  is  an  altitude. 

[5]  A  median  which  is  an  angle-bisector. 

[6]  An  altitude  which  is  an  angle-bisector. 

[7]  A  perpendicular  side-bisector  which  contains  a  vertex.. 

[8]  Two  equal  angle-bisectors. 


CHAPTER   XIII. 

THE   SYMCENTRAL   QUADRILATERAL. 

269.  A  quadrilateral  with  a  symcenter  is  called  a  parallelo- 
gram.   (Ig'm). 

270.  Because  it  has  symcentry,  every 
parallelogram  has  its  opposite  sides  parallel 
and  equal,  its  opposite  angles  equal,  and 
Fig.  127.  ^  diagonals  which  bisect  each  other.  Also, 
■every  straight  through  the  symcenter  cuts  the  parallelogram 
into  congruent  parts. 

271.  Theorem.    A    quadrilateral    with 

'O  \    each    side    parallel    to  its   opposite   is   a 

Fig.  128.  parallelogram. 

Proof.  Since  for  any  two   ||s  the  mid  point  of  the  sect  they 

intercept  on  any  transversal  is  a  symcenter,  .'.  the  mid  point 

of  a  diagonal,  being  a  symcenter  for  both  pairs,  is  a  symcenter 

for  the  quad. 

272.  Theorem.  A  quadrilateral  with  a  pair  of  sides  equal 
and  parallel  is  a  parallelogram. 

Proof.  The  mid  point  of  a  diagonal  is  a  symcenter  for  the 
four  vertices. 

273.  Theorem.  A  quadrilateral  with  each  side  equal  to  its 
c     _!|B_     opposite  is  a  parallelogram. 

Proof.  Any  vertex,  B,  is  the  only  inter- 

\     section  point  of  OA  {^AE]  with  oC\CW\  on 

°      Fig.  129.  that  side  of  the  center-straight  A  C.     But  a 

straight  through  A  \\   to  Z>C  meets  a  straight  through  C  \\  to 

DA  at  that  point,  since  opposite  sides  of  a  ||g'm  are  equal. 

46 


THE   SYMCENTRAL    QUADRILATERAL. 


47 


Fig.  130. 


274.  Theorem.  A  quadrilateral  with  a  pair  of  opposite  sides 
equal  and  each  greater  than  a  diagonal  \b' 
making  equal  alternate  angles  with  the 
other  sides,  is  a  parallelogram. 

Prooif.  For  the  mid  point  C  of  this 
diagonal  is  the  symcenter  of  its  end 
points ;  and  also  of  the  other  two  vertices,  since  one  of  these, 
B,  is  the  one  intersection  point  of  a  semicircle,  whose  center  O 
is  one  end  point  of  this  diagonal,  with  a  ray  starting  in  this 
diameter ;  and  the  other,  B',  is  the  one  intersection  point  of  a 
semicircle  and  ray  symcentral  to  those  with  regard  to  this 
diagonal's  mid  point. 

275.  If  the  sides  given  equal 
were  less  than  the  diagonal  mak- 
ing equal  angles  with  the  other 
sides,  then  the  first  ray  would 
start  from  without  the  first  semi- 
circle and  meet  it  twice  [looking 
at  a  tangent  as  a  secant  through 
two  coincident  points]. 

276.  Theorem.  A  quadrilateral 
with  a  side  equal  to  its  opposite 
and  less  than  a  diagonal  opposite 
€qual  angles  is  a  parallelogram.  '  *^^* 

Proof.  For  the  mid  point  of  the  diagonal  is  the  symcenter 


Fig.  132. 


of  its  end  points  ;  and  also  of  the  other  two  vertices,  since  one 
of  them,  B,  is   the   one  intersection  point  of  an    arc  on  this 


48 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


diagonal  as  chord  with  a  semicircle  whose  center  is  one  end' 
point  of  this  diagonal  and  radius  a  side  less  than  it  ;  and  the 
other  vertex,  B' ,  is  the  one  intersection  point  of  an  arc  and 
semicircle  symcentral  to  those  with  regard  to  this  diagonal's 
mid  point. 

277.  A  circle  OO  {f),  on  a  ray  from  whose  center  a  chord 

00'  is,  can  meet  that  chord  only  once  ; 

but  if  it  cuts  the  arc  of  that  chord  twice 

before    meeting    the   chord,   it    never- 

meets  the  chord. 

278.  If  the  given  equal  sides,  r, 
were  greater  than  the  diagonal  OO'  op- 
posite the  equal  angles,  then  the  first 
semicircle  would  not  meet  the  chord  of 
the  first  arc,  and  so  would  intersect  that 
arc  twice. 

279.  Theorem.  A  quadrilateral  with 
each  angle  equal  to  its  opposite  is  a 
parallelogram. 

Proof.  For  then  any  two  of  the  angles, 
not  opposite  equal  the  other  two,  and  there- 
fore are  supplemental.     So  each  side  is  ||  to 
A  .  . 

Fig.  134.  its  opposite. 

280.  Theorem.  A  quadrilateral- 

^_ \o__Tni^^3'B   whose  diagonals  bisect  each  other 

is  a  parallelogram. 
Fig.  135.  Proof.    Their     intersjection    is 

then  a  symcenter  for  the  four  vertices. 


Fig.  133. 


CHAPTER   XIV. 

SYMMETRICAL   QUADRILATERALS. 

281.  A  symmetrical  quadrilateral  with  a  diagonal  as  axis  is 

called  a  deltoid. 


282.  A  sect  joining  the  mid  points  of  the  opposite  sides  of 
a  quadrilateral  is  called  a  median. 

283.  A  symmetrical  quadrilateral  with  a  median  as  axis  is 
called  a  symtra. 

284.  Theorem.       Every    symmetrical 
quadrilateral  not  a  deltoid  is  a  symtra. 

Proof.  For  to  every  vertex  corresponds 
a  vertex,  hence  the  number  of  vertices  not 
on  the  axis  must  be  even, — here  four  ;  and 
the  sects  joining  corresponding  vertices  are 
bisected  at  right  angles  by  the  axis,  hence  parallel,  hence 
sides ;  for  the  two  diagonals  of  an  undivided  tetragon  can 
never  be  parallel,  since  not  every  pair  of  conjugate  angles  made 

49 


Fig.  137. 


50  ELEMENTARY  SYNTHETIC  GEOMETRY. 

by  the  diagonals  with   the   four  sides  can   be  as  great  as  a 
straight  angle. 

285.  In  any  deltoid,  since  a  diagonal  is  axis  of  symmetry, 
^  therefore  : 

[i]  One   diagonal   [the   axis]   is  the  perpen- 
dicular bisector  of  the  other. 

[2]  One  diagonal  [the  axis]  bisects  the  angles 
at  the  two  vertices. 

[3]  Sides  which  meet  on  one  diagonal  [the 
axis]  are  equal ;  so  each  side  is  equal  to  one  of  its  adjacent 
sides. 

[4]  One  diagonal  [not  the  axis]  joins  the  vertices  of  equal 
angles,  and  makes  equal  angles  with  the  equal  sides. 

[5]  The  triangles  made  by  one  diagonal  [the  axis]  are 
congruent,  and  their  equal  sides  meet. 

[6]  One  diagonal  [not  the  axis]  makes  two  isosceles 
triangles. 

CONDITIONS     SUFFICIENT     TO     MAKE    A    QUADRILATERAL    A 

DELTOID. 

286.  Any  quadrilateral  which  has  one  of  the  six  preceding 
pairs  of  properties  is  a  deltoid  ;  for  from  [i]  that  diagonal  is 
an  axis  of  symmetry ;  from  [2]  that  diagonal  is  axis  ;  from  [3] 
a  AB  =  AD  and  CB  =  CD,  then  the  isosceles  triangles  ABD, 
CBD  have  a  common  axis  of  symmetry,  AC.  This  follows 
also  from  [6]  ;  from  [4]  the  perpendicular  bisector  of  that 
diagonal  must  be  axis  of  symmetry  for  the  two  equal  angles, 
and  their  corresponding  sides  must  intersect  on  it,  hence  it  is 
a  diagonal ;  from  [5]  taking  two  adjacent  sides  equal,  and  the 
angle  contained  by  them  bisected  by  a  diagonal,  then  the  ends 
of  these  equal  sides  are  corresponding  points  with  regard  to 
this  diagonal  as  axis  of  symmetry. 

287.  Theorem.  A  quadrilateral  with  a  diagonal  which  bi- 
sects the  angle  made  by  two  sides,  and  is  less  than  each  of  the 


S  YMME  TRICAL    Q  UADRILA  TERALS. 


51 


Fig.  141. 


other  two  sides,  and  these  sides  equal,  is  a  deltoid  with  this 
diagonal  as  axis. 

Proof.  One  of  the  two  vertices  not  on  this 
diagonal  is  the  one  intersection  point  of  a  semi- 
circle whose  center  is  one  end  point  of  that 
<iiagonal,  with  a  ray  starting  in  its  diameter ; 
and  the  other  is  the  one  intersection  point  of  a 
semicircle,  and  ray  symmetrical  to  those  with  re- 
gard to  this  diagonal. 

288.  Theorem.    A   quadrilateral  with    a  side  meeting  an 
^qual   side    in    a    greater    diagonal    which    is 
opposite  equal  angles  is  a  deltoid  with  that 
-diagonal  as  axis. 

Proof.  Of  the  two  vertices  not  on  this 
diagonal  one  is  the  one  intersection  point  of 
an  arc  on  this  diagonal  as  chord  with  a  semi- 
•circle  whose  center  is  one  end  point  of  this 
•diagonal  and  radius  a  side  less  than  it ;  and  the  other  vertex 
is  the  one  intersection  point  of  an  arc  and  semicircle  sym- 
metrical to  those  with  regard  to  this  diagonal. 

289.  In  any  symtra,  since  a  median  is  axis  of  symmetry, 
therefore  : 

[i]  Two  opposite  sides  are  parallel,  and 
have  a  common  perpendicular  bisector. 

[2]  The  other  two  sides  are  equal,  and 
make  equal  angles  with  the  parallel  sides. 

[3]  Each  angle  is  equal  to  one  and 
supplemental  to  the  other,  of  the  two  not 
opposite  to  it. 

[4]  The  diagonals  are  equal,  and  their  segments  adjacent 
to  the  same  parallel  are  equal. 

[5]  One  median  bisects  the  angle  between  the  two  diago- 
nals, and  also  the  angle  between  the  non-parallel  sides,  when 
produced. 


Fig.  142. 


52 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


CONDITIONS     SUFFICIENT     TO      MAKE     A    QUADRILATERAL    A 

SYMTRA. 
290.  Any  quadrilateral  which  has  one  of   the    preceding^ 
five  pairs  of  properties  is  a  symtra. 

[i]   Here  the  common  perpendicular  bisector  is  an  axis  of 
symmetry. 

[2]  Here  the  perpendicular  bisector  of  the  parallel  sides  is 
a  symmetry  axis  for  the  four  vertices. 

[3]  Since  this  is  the  same  as  two  sides  I  and  the  ^^  adja- 
cent to  either  equal,  therefore  here  the  rt'  bi'  of  the  side 
joining  the  vertices  of  the  equal  angles  is  symmetry-axis  for 
those  vertices  and  angles,  and  for  the  parallel  containing  the 
opposite  side;  .*.  for  the  intersection  points  of  this  parallel 
c  IH  B  with  the  sides  of  the  equal  angles  [the  other 
two  vertices]. 

[4]  Here  since  GB  =  GC,  and  GA  =  GDy 
therefore  the  bi'  of  4.  AGD  is  symmetry-axis 
for  the  four  vertices. 
[5]    Here    since  in  each    of  the    triangles  AGD,   CGB  a 
median  bisects  an  angle,  therefore  it  is  symmetry  axis  for  the 

291.  A    symcentral    deltoid    is    called    a 
rhombus. 

292.  In  a  rhombus : 
[i]  All  four  sides  are  equal. 
[2]   Each  diagonal  is  a  symmetry  axis. 

[3]  Each  diagonal  is  perpendicular  to  the  other,  and  bisects 
two  angles. 

293.  Inversely,  a  quadrilateral  with  [i], 
[2],  or  [3]  is  a  rhombus. 

294.  A   symcentral   symtra  is  called  a 
rectangle. 

295.  In  a  rectangle  : 
\a\  All  its  angles  are  right. 


Fig.  143. 


c 

m' 

B 

-^ 

0,^ 

N 

^\ 

C 

Fig. 

M 

MS. 

A 

S  YMME  TRICAL   Q  UA  DRILA  TERALS. 


53 


[<^]   Each  median  is  a  symmetry  axis. 

[c]  Its  diagonals  are  equal,  and  bisect  each  other. 

296.  Inversely,  a  quadrilateral  with  [ci],  [^], 
or  [c]  is  a  rectangle. 

297.  A  symtral  deltoid  is  called  a  square. 

298.  A  square  has  symcentry,  and  so  has  a  ^ 
rhombus  and  a  rectangle. 

299.  A  quadrilateral  with  [i]  and  [a],  or  [2] 
.and  [U],  or  [3]  and  [c],  is  a  square. 


CHAPTER   XV. 


CONGRUENCE  OF  TRIANGLES, 


300.  Theorem,  Triangles  are  congruent  if  they  have  a  side 
and  two  angles  adjacent  to  it  equal ;  or  a  side  and  two  angles,. 


Fig.  147. 


one  adjacent  and  one  opposite  to  it,  respectively  equal ;  or  two 
sides  and  the  included  angle  equal,  or  two  sides  and  the  angle 
opposite  the  greater  equal,  or  three  sides  equal. 

Proof.  Since  in  any  triangle  the  sum  of  the  three  angles  is 
a  straight  angle,  the  second  case  comes  under  the  first.  In 
every  case,  slide  the  two  triangles  in  the  plane  until  a  pair  of 
equal  sides  coincide,  but  beyond  this  common  side  are  no  coin- 
cident points.  If  then  a  pair  of  equal  angles  have  a  common 
vertex,  or  a  second  pair  of  equal  sides  have  a  common  end 
point,  the  triangles  are  symmetrical  with  regard  to  the  common 
side.     If  not,  symcentral  with  regard  to  its  mid  point. 

54 


CONGRUENCE   OF   TRIANGLES.  55 

301.  Theorem.  If  two  triangles  have  two  sides  and  the 
angle  opposite  the  lesser  equal,  they  either  are  congruent  or 
have  supplemental  angles  opposite  the  greater  equal  sides. 


Proof.  Slide  the  triangles  in  the  plane  until  the  greater 
equal  sides  coincide,  but  beyond  this  common  side  are  no  co- 
incident points.  Then  in  one  triangle  the  vertex  opposite  the 
common  side  is  one  of  the  two  intersection  points  of  a  secant 
from  one  end  point  with  a  semicircle  whose  center  is  the  other 
end  point  of  the  common  side,  and  in  the  other  triangle  is  one 
of  two  points,  which,  if  the  angles  given  equal  have  now  a 
common  vertex,  are  symmetrical  to  these  with  regard  to  the 
common  side ;  if  not,  symcentral  with  regard  to  its  mid  point. 
If  corresponding  points  of  these  four  be  vertices,  the  triangles 
are  congruent.  If  not,  then  opposite  the  common  side  the 
angle  in  one  triangle  equals  the  exterior  angle  in  the  other. 

302.  Corollary.  If  two  triangles  have  two  sides  of  the  onq 
equal  respectively  to  the  sides  of  the  other,  and  the  angles 
opposite  to  one  pair  of  equal  sides  equal,  then,  if  the  angles 
opposite  the  other  pair  of  equal  sides  are  not  supplemental,  or 
if  any  one  angle  in  either  triangle  is  a  right  angle,  the  triangles 
are  congruent. 


56 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


t^       


303.  Theorem.   Of  sects  joining  two 
A  symmetrical   points  to  a  third,  that  cut- 
ting the  axis  is  the  greater. 
Proof.  BA  =  BC-\-  CA'; 
£C+  CA'  >  BA'. 

304.  Theorem.  If  two  triangles  have  two 
sides  of  the  one  respectively  equal  to  two  sides 
of  the  other,  but  the  included  angles  unequal, 
then  that  third  side  is  the  greater  which  is  op- 
posite the  greater  angle. 

Proof.  Slide  the  triangles  in  the  plane  until 

a  pair  of  equal   sides   coincide,  and  the  other 

pair  of  equal  sides  have  a  common  end  point. 

Bisect  the  angle   made  by  these  equal  sides. 

This  axis  cuts  the  third  side  which  is  opposite 

the  greater  angle. 

\  305.  Inverse.  If  two  triangles  have  two  sides 

Fig.  150a.  qJ  ^YiQ  one  respectively  equal  to   two   sides  of 

the  other,  but  the  third  sides  unequal,  then,   of  the  angles 

opposite  these  third  sides,  that  is  the  greater  which  is  opposite 

the  greater  third  side. 

306.  Theorem.  If  three  parallels  intercept  equal  sects  on 
one  transversal  they  intercept  equal  sects 
on  every  transversal. 

Proof.  If,  on  a  straight,  AB  =  BC, 
and  f  through  A,  B,  and  C  intersect 
another  st'  in  A',  B\  C,  then  st's  through 
B'  and  C,  drawn  j|  to  BC,  make  ^  A^ 
307.  Corollary  I.  The  intercept  made 
on  the  mid  of  three  parallels  by  two  transversals  differs  from 
the  intercepts  on  the  others  by  equal  sects. 

308.  Cor.  II.  If  through  the  mid  point  of  one  side  of  a 
triangle  a  straight  be  drawn  parallel  to  a  second  side,  it  will 
bisect  the  third  side. 


Fig.  151. 


CONGRUENCE   OF   TRIANGLES. 


57 


309.  Cor.  III.  If  a  straight  parallel  to  one  side  of  a  trian- 
-gle  cuts  off  any  fractional  part  of  a  side,  it  cuts  off  the  same 
fraction  of  the  other  side. 

310.  Inverse.  The  sect  joining  the  mid  points  of  any  two 
sides  of  a  triangle  is  parallel  to  the  third  side,  and  equal  to  half 
of  it. 

311.  Corollary.  The  sect  joining  points  which  bound  with 
any  vertex  of  a  triangle  the  same  fractional  parts  of  two  sides 
is  parallel  to  the  third  side  and  is  that  fractional  part  of  it. 


ROTATION-CENTER. 


312.  Theorem.  In   a  plane,  the  result  of  sliding  any  poly- 
gon is  the  same  as  of  a  rotation  about  a  fixed  point. 


Fig.  152. 

Proof.  Join  vertex  A'  with  its  trace  A'\  and  B'  with  B". 
The  perpendicular  bisectors  of  A' A"  and  B' B"  intersect  in  the 
rotation-center  O.  For  A  A' OB'  ^  A  A" OB"  [having  three 
sides  respectively  equal]. 

Consequently  t  A'OA"  =  t  B'OB". 

313.  The  altitudes  of   a  triangle  are  l  b 

concurrent,  and  the  point  is  called  the  \ 
triangle*s  orthocenter. 

They  must  cointersect,  since  each  con- 
tains the  circumcenter  of  a  triangle  made 
by  drawing  through  the  vertices  of  the 
given  triangle  parallels  to  its  sides.  Fig.  153. 


58  ELEMENTARY  SYNTHETIC  GEOMETRY. 


EXERCISES  ON  BOOK   I. 

1.  If,  with  the  vertex  of  an  angle  as  center,  two  circles  be  described, 
and  the  points  in  which  they  cut  its  arms  be  joined,  the  joins  are  ||  or 
intersect  on  the  angle's  bisector. 

2.  In  a  -I-  A  two  sides,  two  altitudes,  two  medians,  two  -^-bisectors 
are  =,  and  cross  on  the  axis. 

3.  Intersecting  equal  circles  are  -I-  with  regard  to  their  common 
chord. 

4.  If  about  two  given  points  as  centers  pairs  of  equal  intersecting  Gs 
be  described,  all  the  pairs  have  their  common  points  on  one  straight. 

5.  If  of  two  convex  polygons  one  is  wholly  within  the  other,  then  the 
outer  has  the  greater  perimeter. 

6.  An  interior  angle  of  a  regular  dodecagon  is  what  fraction  of  a 
r't  4.  ? 

7.  If  two  sides  of  a  A  be  produced  through  their  common  vertex 
until  each  is  doubled,  the  join  of  the  ends  is  ||  to  the  third  side. 

8.  The  join  of  the  points  of  contact  of  ||  tangents  to  a  O  is  a  diam- 
eter. 

9.  If  in  a  -I-  A  we  decrease  one  of  the  equal  sides  and  increase  the 
other  equally,  the  join  of  the  points  so  obtained  is  bisected  by  the  third 
side. 

10.  In  A  ABC,  if  r't  bi'  oi  a  cuts  the  st'  d  in  D  and  c  in  E,  then  is 

4  ABD  =  4-  ACE  =  dif  between  ^s  B  and  C. 

11.  If  from  two  p'ts  of  a  st',  ±s  to  another  st'  are  =,  either  the  st's 
are  ||,  or  the  sects  from  their  cross  to  the  p'ts  are  =. 

12.  The  sum  of  the  ±s  to  the  =  sides  from  any  p't  in  the  third  side 
of  a  -I-  A  equals  one  of  the  =  altitudes. 

13.  All  equal  sects  between  two  ||s  belong  to  two  sets  of  \\s. 

14.  If  from  the  vertices  in  the  same  sense  on  the  sides  of  a  ]|g'm  a 
given  sect  be  taken,  the  points  so  obtained  are  vertices  of  a  ||g'm  cosym- 
central  with  the  first. 

15.  Find  the  bisector  of  an  4  without  using  its  vertex. 

16.  A  quad'  with  two  sides  ||  and  the  others  =  is  either  a  ||g'm  or  a 
symtra. 

17.  If  two  sides  of  a  quad'  have  a  common  r't  bi',  it  is  a  symtra. 

18.  The  r't  bi's  of  the  non-||  sides  of  a  symtra  cross  on  the  r't  bi'  of 
the  other  sides. 


EXERCISES  ON  BOOK  I.  59 

19.  If  the  diagonals  of  a  quad'  are  =,  its  medians  are  ±. 

20.  If  in  a  trapezoid  three  sides  are  =,  then  the  angles  adjacent  to 
the  fourth  side  are  bisected  by  the  diagonals. 

21.  The  sects  to  the  intersection  points  of  a  secant  from  the  ±  pro- 
jections of  ends  of  a  diameter  on  it  are  =. 

22.  A  quad'  is  fixed  by  5  given  magnitudes. 

23.  An  «-gon  is  fixed  by  2«  —  3  given  magnitudes. 

24.  The  bisector  of  an  ^  of  a  a  and  the  r't  bi'  of  the  opposite  side 
cross  on  the  circum-O. 

25.  The  cross  of  an  altitude  (produced  through  its  foot)  with  the  cir- 
cum-0  is  -I-  to  the  orthocenter  with  respect  to  that  side  of  the  A. 

26.  Whether  their  vertex  be  on  or  within  the  O,  a  pair  of  vertical 
angles  together  intercept  the  same  part  of  a  O. 

27.  Vertical  r't  ^s  with  vertex  on  or  within  a  0  intercept  half  of  it. 

28.  Joining  one  common  p't  of  two  =  intersecting  Os  to  the  crosses, 
of  a  secant  through  the  other  common  point  gives  =  sects. 

29.  If  from  one  intersection  p't  of  two  =  Os  as  center  we  describe 
any  third  circle  cutting  them,  then  the  four  intersection  p'ts  are  vertices 
of  a  symtra  whose  non-||  sides  go  through  the  other  intersection  p't  of 
the  =  Os. 

30.  A  0  on  the  common  chord  of  .two  =  Os  as  diameter  bisects  all 
sects  through  an  intersection  p't  of  the  Os  and  ending  in  them. 

31.  A  symtra  is  cyclic. 

32.  A  deltoid  is  a  circumscribed  quad'. 

33.  The  four  ^-bisectors  of  a  quad'  make  a  cyclic  quad'. 

34.  The  four  crosses  of  the  inner  with  the  outer  common  tangents  to- 
two  Os  lie  on  a  circle  with  their  center-sect  as  diameter, 

35.  The  sect  of  an  outer  between  the  inner  tangents  equals  the  sect 
of  an  inner  between  its  points  of  contact. 

36.  Each  side  of  a  A  is,  by  the  p'ts  of  contact  of  the  in-Oand  an  ex-O. 
divided  into  three  sects,  of  which  the  outer  two  are  =. 

37.  If  a  polygon  has  a  circum-O  and  a  concentric  in-O,  it  is  regular. 

38.  To  make  a  regular  hexagon,  trisect  the  sides  of  a  regular  trigon 
and  join  the  points  next  its  vertices. 

39.  To  make  a  regular  octagon,  about  each  vertex  of  a  square,  with 
half  the  diagonal  as  radius,  describe  a  O  and  join  the  crosses  next  its 
vertices. 

40.  If  a  p't  of  its  circum-O  be  joined  to  the  vertices  of  a  regular  A^ 
the  greatest  sect  equals  the  sum  of  the  other  two. 


BOOK   11. 

PURE  SPHERICS. 


CHAPTER   I. 

PRIMARY  CONCEPTS. 

314.  A  circle  is  a  closed  line  that  will  slide  in  its  trace. 
Though  in  itself  unbounded  and  everywhere  alike,  yet  it  is 
finite.  On  it  two  points  starting  from  coincidence  and  moving 
in  opposite  senses  will  meet. 

315.  Every  point  on  a  circle  has  one  other  on  it  such  that 
the  two  bisect  the  circle.     Two  such  are  called  opposite  points. 

316.  If  a  pair  of  opposite  points  can  be  kept  fixed  while  a 
circle  moves,  it  describes  a  surface  called  a  sphere. 

317.  A  sphere  is  a  closed  surface  which  will  slide  in  its 
trace.  Though  in  itself  unbounded  and  everywhere  alike,  yet 
it  is  finite,  being  generated  completely  by  one  finite  motion  of 
a  finite  line. 

318.  Assumption  I.  Any  figure  drawn  on  the  sphere  may 
be  moved  about  in  the  sphere  without  any  other  change. 

319.  Assumed  Construction  I.  Through  any  two  points,  in 
a  sphere,  can  be  passed  a  line  congruent  with  the  generating^ 
line  of  the  sphere. 

In  Book  II.  g-line  will  always  mean  such  a  line,  and  sect 
will  mean  a  piece  of  it  less  than  half. 

60 


PRIMARY    CONCEPTS. 


61 


320.  Assumption  II.  Two  sects  cannot  meet  twice  on  the 
sphere. 

If  two  sects  have  two  points  in  common,  their  g-Iines  coin- 
cide throughout.  Through  two  points,  not  opposite  points  of 
a  g-hne,  only  one  distinct  g-Hne  can  pass. 

321.  A  piece  of  the  sphere  with  part  of  a  g-Hne  as  one  of 
its  boundaries,  would  fit  all  along  the  g-line  from  either 
side. 

322.  Because  of  the  symmetry  in  its  generation,  the  sphere 
is  cut  by  any  g-line  on  it  into  two  equal  parts,  called  hemi^ 
spheres.  ' 

323.  If  one  end  point  of  a  sect  be  kept 
fixed,  the  other  end  point  moving  in  the 
sphere  describes  what  is  called  an  arc,  and 
the  sect  is  said  to  rotate  in  the  sphere  about 
the  fixed  end  point.  The  arc  is  greater  as 
the  amount  of  rotation  is  greater. 

324.  Two  sects  from  the  same  point, 
when  looked  at  with  special  reference  to 
the  amount  of  rotation  necessary  to  bring 
their  g-lines  into  coincidence,  are  said  to 
form  a  spherical  angle.  The  spherical 
angle  is  greater  as  the  amount  of  rotation 
is  greater. 

325.  When  a  sect  has  rotated  just  suf- 
ficiently to  fall  again  into  the  same  g-line, 
the  angle  described    is  called  a  straight    angle,  and    the  arc 
described  is  called  a  semicircle. 

326.  Half  a  straight  angle  is  called  a  right  angle. 

327.  The  whole  angle  about  a  point  in  the  sphere,  that  is, 
the  angle  described  by  a  sect  rotating  until  it  coincides  with  its 
trace,  is  called  a  perigon ;  the  whole  arc  is  called  a  circle. 


Fig.  155. 


62 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


The  fixed  end  point  is  called  a  pole  of  the  circle,  and  the 
sect  is  called  a  spherical  radius  of  the  circle. 

328.  Assumed  Construction   II.  A  circle  can   be    described 
from  any  pole,  with  any  sect  as  radius. 

329.  Assumption  III.  All  straight  angles  are  equal. 

330.  Corollary  I.  All  perigons  are  equal. 


331.  Corollary  II.  The  two  angles  on 
the  same  side  of  a  g-line,  made  by  a  sect 
with  one  extremity  in  that  g-line,  are 
together  k  straight  angle. 

332.  Corollary  III.  Vertical  angles  are 
equal,  being  supplements  of  the  same 
angle. 

333.  Theorem.  Every  g-line  in  the 
sphere  cuts  every  other  in  two  opposite 
points. 

Proof.  Let  BB'  and  CC  be  any  two 
g-lines.  Since  each  bisects  the  sphere, 
therefore  the  second  cannot  lie  wholly 
in  one  of  the  hemispheres  made  by  the 
first,    therefore    they   intersect  at   two 


Fig.  157. 


points,  which  are  therefore  opposite. 


334.  On  a  sphere,  every  circle  has  two 
poles,  which  are  opposite  points,  and  its« 
spherical  radius  to  one  pole  is  the  supple- 
ment of  that  to  the  other. 

335.  A  spherical  figure  made  by  two 
half  g-lines  intersecting  in  opposite  points, 
is  called  a  lune. 


PRIMARY    CONCEPTS. 


63 


336.  Theorem.  The  angle  contained  by  the  sides  of  a  lune 
at  one  of  their  points  of  intersection  equals 
the  angle  contained  at  the  other. 

Proof.  Slide  the  lune,  in  the  sphere, 
until  each  of  the  two  intersection  points 
falls  on  the  trace  of  the  other,  and  one  of 
the  half  g-lines  on  the  trace  of  the  other. 
If  the  angles  were  unequal,  the  smaller 
could  thus  be  brought  within  the  trace  of 
the  greater,  and  its  second  half  g-line 
would  start  between  the  traces,  and  since 
it  could  meet  neither  again  until  it  reached  the  opposite  inter- 
section point,  we  would  find  the  surface  of  the  lune  less  than 
its  trace. 

337.  One  quarter  of  a  g-line  is  called  a  quadrant. 


338.  A  spherical  polygon  is  a  closed 
figure,  in  the  sphere,  bounded  by  sects, 
no  two  of  which  cross. 

339.  A  spherical  triangle  is  a  three- 
sided  spherical  polygon,  with  no  interior 
angle  greater  than  a  straight  angle. 


340.  A  spherical  triangle  is  positive 
[-(-]  if  a  sect  with  one  end  pivoted  within 
it  and  rotating  counter-clockwise,  after 
passing  through  the  vertex  of  the  greatest 
angle  goes  next  over  the  vertex  of  the 
least. 


Fig.  161. 


CHAPTER   II. 


SYMCENTRY    ON   THE   SPHERE. 


341.  On  a  sphere  a  point  has,  with  regard  to  a  given  sym- 
center,  always  one  and  only  one  symcentral  point,  namely,  the 
one  which  ends  the  sect  from  the  given  point  bisected  by  the 
symcenter. 

342.  Two  figures  are  symcentral  when 
they  can  be  placed  so  as  to  have  a  sym- 
center. 

One  figure  is  symcentral  when  it  has  a 
symcenter,  that  is,  a  point  with  respect  to 
which  every  point  of  the  figure  has  its- 
symcentral  point  on  the  figure. 

A  lune  is  symcentral  with  regard    to 
the  cross  of  the  g-line  bisecting  its  angles, 
with  the  g-line  bisecting  its  sides. 
343.  Symcentral  figures  on  a  sphere  have  precisely  the  same 
properties  as  in  the  plane,  including  congruence. 

64 


Fig.  162. 


CHAPTER    III. 

SYMMETRY   ON  THE  SPHERE. 

344.  Two  points  on  a  sphere  are  symmetrical  with  respect 
to  a  g-line,  when  it  bisects  at  right  angles 
the  sect  joining  them. 

This  g-line  is  called  their  axis  of  sym- 
metry. 

345.  Two  points  on  a  sphere  have 
always  one,  and  only  one,  symmetry  axis 
on  that  sphere. 

346.  A  point  has,  with  regard  to  a 
given  axis  of  symmetry,  always  one,  and 
only  one,  symmetrical  point,  namely,  the  one  which  ends  the 
sect  from  the  given  point  perpendicular  to  the  axis  and 
bisected  by  the  axis. 

347.  Two  figures  on  the  sphere  have  an  axis  of  symmetry 
when,  with   regard  to    this   g-line,   every 
point   of  each   has  its  symmetrical  point 
on  the  other. 

348.  Two  figures  are  symmetrical  when 
they  can  be  placed  so  as  to  have  an  axis 
of  symmetry. 

349.  One  spherical  figure  has  an  axis  of 
symmetry  when,  with  regard  to  this  g-line, 
every  point  of  the  figure  has  its  sym- 
metrical point  on  the  figure. 

350.  One  figure  is  symmetrical  when  it  has  an  axis  of  sym- 
metry. 

351.  Any  figure  on  the  sphere  has,  with  regard  to  any  g-line 
on  the  sphere  as  axis,  always  one,  and  only  one,  symmetrical 
figure. 

65 


Fig.  163. 


66 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


352.  One  figure  is  symmetrical  when  it 
has  an  axis  with  regard  to  which  its  sym- 
metrical figure  coincides  with  itself. 

353.  Every  point  on  an  axis  is  sym- 
metrical to  itself. 

Fig.  164, 

354.  Assumption  IV.  The  figure  sym- 
metrical to  a  sect  is  an  equal  sect ;  to  a 
spherical  angle,  is  a  spherical  angle  equal 
in  magnitude  but  opposite  in  sense. 

355.  Corollary.  A  sect,  or  g-line,  or 
spherical  angle,  in  one  of  two  symmetrical 
figures,  has  a  symmetrical  sect,  or  g-line, 

'°'  *  ^'  or  spherical  angle,  in  the  other. 

356.  The  intersection  point  of  two  sects  is  symmetrical  to 
the  intersection  of  two  symmetrical  to  those. 

357.  The  intersection  points  of  two  symmetrical  g-lines  are 
on  the  axis. 

358.  The  bisector  of  a  spherical  angle  is  symmetrical  to 
the  bisector  of  the  symmetrical  spherical  angle. 

359.  The  angle  between  two  symmetrical  g-lines  is  bisected 
iby  the  axis. 

360.  Two  g-lines  are  symmetrical  with  regard  to. either  of 
their  angle-bisectors. 

For  there  is  a  g-llne  symmetrical  to 
the  first  with  regard  to  that  angle-bisector, 
and  the  angle  between  the  two  symmetri- 
cal g-lines  is  bisected  by  the  axis. 

361.  Any  g-line  is  symmetrical  with 
regard  to  any  of  its  perpendiculars. 

362.  Any  circle    is   symmetrical  with 
Fig.  166.               regard  to  any  of  its  spherical  diameters. 


SYMMETRY  ON   THE   SPHERE. 


67 


363.  Every  point  on  the  perpendicular  bisector  of  a  sect  is 
the  pole  of  a  circle  passing  through  its  end 
points. 

For  A-\'B\  axis  CD ;  .-.  CA  =  CB. 

Thus  sects  from  any  point  on  its  per- 
pendicular bisector  to  the  end  points  of 
the  sect  are  equal. 

364.  The  perpendicular  bisector  of  a 
spherical  chord  contains  the  poles  of  the 
circle.     For  the  end  points  of  the  chord  ^'<^-  ^^^• 

are  symmetrical  with   regard   to  this  perpendicular,  and  also 
with  regard  to  the  perpendicular  from  a  pole. 

365.  Two  circles  with  three  points  in  common  coincide. 

366.  One  spherical   radius,  of  every  circle  on  the  sphere,  is 
less  than  a  quadrant. 

Call  its  pole  the  q-pole,  and  it  the  q-radius. 

367.  If  the  q-pole-sect  of  two  circles 
equals  the  sum  of  their  q-radii,  they  have  a 
common  point  on  their  q-pole-sect,  and  by 
symmetry  no  other  common  point. 

Such  circles  are  said  to  be  tangent  ex- 
ternally. 

Neither  has  a  point  in  common  with 
a  circle  concentric  with  the  other,  but  of 
lesser  q-radius. 

368.  If  the  q-pole-sect  equals  the  difference  of  the  q-radii, 
the    two   circles   have  a  common    point  on 
their  pole-g-line,  and  by  symmetry,  no  other 
common  point. 

Such  circles  are  said  to  be  tangent  in- 
ternally. 

Neither  has  a  point  in  common  with  a 
circle  concentric  with  the  lesser  and  of  lesser 
q-radius. 


Fig.  168. 


Fig.  169. 


68  ELEMENTARY  SYNTHETIC  GEOMETRY. 

369.  While  the  q-pole-sect  is  growing,  from  equality  with 
the  difference  of  the  q-radii,  up  to  equality  with  their  sum,  the 
two  circles  have  always  two  common  points,  symmetrical  with 
regard  to  their  pole-g-line. 

370.  Problem.  To  make  a  spherical  triangle,  given  its  sides. 
Construction.  If    two    of   its   sides   are    each    less    than    a 

quadrant,  then  with  these  as  q-radii,  and  the  end  points  of  the 
third  side  as  poles,  describe  two  circles.  Their  two  common 
points  will  be  the  third  vertices  of  two  symmetrical  triangles 
with  the  three  given  sides. 

If  two  of  the  given  sides  are  each  greater  than  a  quadrant, 
take,  in  the  above,  their  supplements  with  the  given  third  side. 
Then  in  the  two  triangles  obtained,  produce  these  two  supple- 
ments until  they  meet. 

These  two  meeting  points  will  be  the  third  vertices  of  two 
symmetrical  triangles  with  the  three  given  sides. 

371.  Corollary  I.  Any  two  sides  of  a  spherical  triangle  are 
together  greater  than  the  third. 

For  if  two  be  each  less  than  a  quadrant,  and  together  equal 
to  the  third,  the  construction  circles  will  be  tangent  extenally. 

If  two  be  each  greater  than  a  quadrant,  their  difference  is 
that  of  their  supplements,  which  is  less  than  the  third  side  ;  for 
if  equal  to  it,  the  construction  circles  would  be  tangent  inter- 
nally. 

372.  Corollary  II.  The  sum  of  the 
three  sides  of  any  spherical  triangle  is  less 
than  a  g-line. 

373.  Since  any  chord  is  bisected  by  the 
perpendicular  from  a  pole,  .*.  a  g-line  J_ 
to  a  diameter  at  an  end  point  has  only 
this  point  in  common  with  the  circle. 

Fig.  170.  This  point  of  the  circle  is  symmetrical 

to  itself  with  regard  to  this  diameter  as  axis. 

But  if  we  draw  through  this  point  B  any  g-line  BF  not_L  to 


SYMMETRY  ON    THE   SPHERE. 


69 


the  spherical  radius  AB,  then  the  perpendicular  from  a  pole  A 
will  meet  this  g-hne  BF  at  some  other  point  E. 

Hence  the  g-line  BF  cuts  the  circle  again  at  B'  •{•  B,  axis 
AE;  .-. 

Theorem.  At  every  point  on  the  circle  one,  and  only  one, 
tangent  can  be  drawn,  namely,  the  perpendicular  to  a  radius  at 
that  point. 

374.  Let  P  be  a  point  not  in  the  g-line  g, 
2iX\dPC^_tog  :  then^  is  tangent  to  0P[/'C] 
at  C. 

If  PC  is  less  than  a  quadrant,  any  second 
circle  with  q-radius  <  PC,  and  q-pole  P,  lies 
wholly  within  ©  P  [_PC\     Therefore: 

Theorem.  If  less  than  a  quadrant,  the 
perpendicular  is  the  least  sect  between  a 
point  and  a  g-line. 

375.  The  poles  of  all  circles  tangent  to 
two  intersecting  g-lines  are  in  their  angle- 
bisectors. 

376.  From  any  point  on  an  angle-bisec- 
tor the  perpendiculars  to  the  g-lines  are 
equal. 

377.  Theorem.  If  two  angles  of  a 
triangle  be  equal,  the  triangles  is  isos- 
celes. 

Proof.  The  perpendicular  bisector  of 
the  side  joining  the  equal  angles  is  the 
symmetry  axis  for  that  side  and  its  end 
points,  and  so  for  angles  made  with  that 
side  at  those  points  which  are  equal  in 
magnitude  and  opposite  in  sense. 

378.  Theorem.  If  one  angle  of  a  spherical  triangle  is  greater 
than  a  second,  the  side  opposite  the  first  must  be  greater 
than  the  side  opposite  the  second. 


Fig.  17a. 


Fig.  173. 


70 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


Fig.  174. 


Proof.  Given  the  4.C>  4.  A. 

If  i.  DCA  =  4A,  then  DC=DA. 

But  DC+  DB  >  BC;  .-.  DA  +  DB 
>  BC. 

379.  Inverse.  If  one  side  of  a  spheri- 
cal triangle  is  greater  than  a  second,  the 
angle  opposite  the  first  must  be  greater 
than  the  angle  opposite  the  second. 

Proof.  For  the  angle  opposite  the 
second  cannot  be  the  greater,  nor  can 
they  be  equal. 

380.  Theorem.  In  an  isosceles  triangle  the  angles  opposite 
the  equal  sides  are  equal. 

Proof.  The  bisector  of  the  angle  between  the  equal  sides 
is  a  symmetry  axis  for  those  sides  and  their  end  points,  hence 
for  the  triangle. 

381.  Corollary.  In  an  isosceles  triangle  the  bisector  of  the 
angle  between  the  equal  sides  is  perpendicular  to  the  third 
side. 

382.  If  the  vertices  of  a  polygon  are 
concyclic,  the  polygon  may  be  called 
cyclic. 

383.  In  a  cyclic  quadrilateral,  the  sum 
of  one  pair  of  opposite  angles  equals 
the  sum  of  the  other  pair. 

Proof.  Join  the  circumcenter  E  with 
A,  B,  C,  D,  the  vertices.     By  isosceles  tri- 
FiG.  X7S.  angles,  ^  ABC  =  ^  BAE -\-  ^  ECB,  and 

CDA  =  t  DCE  +  ^  DAE. 


CHAPTER   IV. 


THE  SYMCENTRAL  QUADRILATERAL. 


diagonal  making 


384.  A  symcentral  spherical  quadrilateral,  or  cenquad,  has 
its  opposite  sides  equal,  its  opposite  angles  equal,  and  diagonals 
which  bisect  each  other. 

Also,  every  g-line  through  the  symcenter  cuts  the  cenquad 
into  congruent  parts. 

385.  Theorem.  A  quadrilateral  with  a 
with  each  side  an  angle  equal  to  its  alternate, 
is.  a  cenquad. 

Proof.  The  mid  point  of  this  diagonal  is 
a  symcenter  for  both  pairs  of  opposite  sides. 

386.  Theorem.  A  quadrilateral  with  a 
pair  of  opposite  sides  equal  and  making 
equal  alternate  angles  with  a  diagonal,  is  a 
cenquad. 

Proof.  The  mid  point  of  the  diagonal  is  a  symcenter  for 
the  four  vertices. 

387.  Theorem.  A  quadrilateral  with  a  pair  of  opposite  sides 
equal,  and  a  diagonal  making  equal  alternate  angles  with  the 
other  sides  and  opposite  angles  not  supplemental,  is  a  cenquad. 

Proof.  The  mid  point  of  this  diagonal  is  the  symcenter  of 
its  end  points;  and  also  of  the  other  two  vertices,  since  one  of 
these  is  an  intersection  point  of'a  semicircle,  of  which  a  diame- 
ter is  bisected  by  one  end  point  of  this  diagonal,  with  a  g-line 
through  its  other  end ;  and  the  other  is  the  symcentral  inter- 

71 


Fig.  176. 


72  ELEMENTARY  SYNTHETIC  GEOMETRY. 

section  point   of  a  semicircle  and  g-line  symcentral  to  those 
with  regard  to  this  diagonal's  mid  point. 

388.  Theorem.  A  quadrilateral  with  each 
side  equal  to  its  opposite  is  a  cenquad. 

Proof.  Any  vertex,  B,  is  the  only  inter- 
section point  of  QA  [AB']  with  OC  [CB] 
on  that  side  of  their  pole-g-line,  AC. 

But  the  fourth  vertex  of  a  cenquad  with 
Fig.  177.  sides  CD  =  AB  and  DA  =  CB,  and  symcen- 

ter  the  mid  point  of  AC,  is  that  point  B, 

389.  Theorem.  A  quadrilateral  whose  diagonals  bisect  each 
other  is  a  cenquad. 

Proof.  Their  intersection  is  then  a  symcenter  for  the  four 
vertices. 


CHAPTER  V. 


SPHERICAL  TRIANGLES. 


390.  Theorem.  Spherical  triangles  of  the  same  sense  are 
congruent  if  they  have  a  side  and  two  angles  adjacent  to  it 
equal ;  or  two  sides  and  the  included 
angle  equal ;  or  two  sides  and  the  angles 
opposite  one  pair  equal,  opposite  the 
other  pair  not  supplemental;  or  three 
sides  equal. 

Proof.  Slide  the  two  triangles  in  the 
sphere  until  a  pair  of  equal  sides  coincide, 
but  beyond  this  common  side  are  no  co- 
incident points.     The  triangles  are  then  ^'°-  ^'''^• 
symcentral  with  regard  to  the  mid  point  of  the  common  side. 

391.  Triangles  which  would  be  con- 
gruent, but  that  they  differ  in  sense,  are 
symmetrical.  Symmetrical  triangles  are 
of  opposite  sign. 

392.  Corollary.  Symmetrical  isosceles 
spherical  triangles  are  congruent. 

For  the  equality  of  two  angles  in  a  tri- 
angle obliterates  the  distinction  of  sense 
or  sign. 

73 


Fig.  i8o. 


74  ELEMENTARY  SYNTHETIC  GEOMETRY. 

393.  Theorem.  An  exterior  angle  of  a 
spherical  triangle  is  greater  than,  equal  to, 
or  less  than  either  of  the  interior  opposite 
angles,  according  as  the  median  from  the 
other  interior  opposite  angle  is  less  than, 
equal  to,  or  greater  than  a  quadrant. 

Proof.  Let  A  CD  be  an  exterior  angle 
"f^^T"^  of  the    A  ABC.     Bisect  AC  dit  E.     Join. 

BE,  and  produce  to  F,  making  EF  ^=  BE.     Join  EC. 

2  ABE  ^  2  CFE. 

[Spherical  triangles  of  the  same  sense  having  two  sides- 
and  the  included  angle  equal  are  congruent.] 

.-.  4.BAE  =  ^FCE. 

If,  now,  the  median  BE  be  a  quadrant,  BEF  is  a  half-g-line, 
and  F  lies  on  BD;  .-.  4.DCE  coincides  with  :^FCE^ 
.'.  tDCE  =  4.BAE. 

If  the  median  BE  be  less  than  a  quadrant,  BEF  is  less  than 
a  half-g-Hne,  and  F  lies  between  CD  and  AC\  .'.  ^DCA  > 
^FCE,  .-.  DCA  >  ^BAC 

And  if  BE  be  greater  than  a  quadrant,  BEF  is  greater  than 
a  half-g-line,  and  F  lies  between  CD  and  AC  produced 
through  C;  .:   ^DCA  <  4.FCE,  .:  DCA  <  ^BAC. 

Thus,  according  as  BE  is  greater  than,  equal  to,  or  less 
than  a  quadrant,  the  exterior  ^ACD  is  less  than,  equal  to,  or 
greater  than,  the  interior  opposite  ^BAC. 

394.  Inversely,  according  as  the  exterior  angle  A  CD  is 
greater  than,  equal  to,  or  less  than  the  interior  opposite  angle 
BAC,  the  median  BE  is  less  than,  equal  to,  or  greater  than  a 
quadrant. 


SPHERICAL    TRIANGLES. 


n 


Fig.  182. 


395.  Theorem.  Any  two  perpendiculars  to  a  g-line  intersect 
in  two  points,   from   either  of  which  all 
sects  drawn  to  that  g-line  are  quadrants 
perpendicular  to  it. 

Proof.  Let  AB  and  CB,  drawn  at  right 
angles  to  AC,  intersect  at  B,  and  meet  AC 
again  at  A'  and  C,  respectively. 

Then  ^BA'C'=^BAC'  and  i^BCA' 
=  ^BCA'. 

[The  angles  contained  by  the  sides  of 
a  lune,  at  their  two  points  of  intersection,  are  equal.] 

Moreover,  AC  =  AC,  for  they  have  the  common  supple- 
ment A  C.  Hence,  keeping  A  and  C  on  the  line  A  C,  slide  ABC 
until  ^6"  comes  into  coincidence  with  A'C.  Then  the  angles 
2itA,C,A',  C  being  all  right,  AB  will  lie  along  A'B,  and  CB 
along  CB,  and  hence  the  figures  ABC  2iV\d  A' BC  coincide. 

Therefore  each  of  the  half-lines  y^^^'  and  CBC  is  bisected 
at^. 

In  like  manner,  any  other  line  drawn  at  right  angles  to  AC 
passes  through  B,  the  mid  point  of  ABA'. 

Hence  every  sect  from  AC  to  B  \s3i  quadrant  ^  to  AC. 

396.  Corollary  I.  A  g-line  is  a  circle  whose  spherical  radius 
is  a  quadrant. 

397.  Corollary  H.  A  point  which  is  a  quadrant  from  two 
points  in  a  g-line,  and  not  in  the  g-line,  is  its  pole. 

398.  Corollary  HI.  Equal  angles  at  the  poles  of  lines  inter- 
cept equal  sects  on  those  lines. 

399.  T^iQ polar  of  any  point  is  the  g-line  of  which  that  point 
is  a  pole. 

400.  If  an  angle  be  a  fraction  of  a  perigon,  it  intercepts  on 
the  polar  of  its  vertex  that  fraction  of  a  g-line. 

401.  Theorem.  If  a  median  is  a  quadrant,  it  is  an  angle- 
bisector,  and  the  sides  of  the  bisected  angle  are  supple- 
mental. 


v^ 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


Proof.  The  quadrant  and  the  sides  BA,  BC,  all  produced, 
are  concurrent  in  B'  opposite  to  B. 
.'.  ABCB'A,  is  a  cenquad  [its  diagonals  AC 
and  BB'  bisect  each  other].  .-.  AB=CB', 
and  AB,  BC  are  supplemental.  Also,  the 
complements  of  AB  and  CB'  are  equal, 
i.e.,  AH ^^  CF \  also,  the  supplements 
of  4.  BAD  and  ^  BCD,  or  t  DAH  = 
^DCF;  .'.  2AD//^  a  CD  F[t\vo  sides 
and  included  angle];  .-.  ND  =  DF;  .-. 
^  ABC  is  bisected  by  BD. 

402.  Corollary.  If  two  sides  of  a  triangle  are  supplemental, 
the  opposite  angles  are  supplemental. 

403.  Theorem.  Two  spherical  triangles,  of  the  same  sense, 
having  two  angles  of  the  one  equal  to  two  angles  of  the  other, 
•the  sides  opposite  one  pair  of  equal  angles  equal,  and  those 

opposite  the  other  pair  not  supplemental, 
are  congruent. 

Proof.  Given  4B=:^E;  4- C  —  F ; 
AB  =  DE;  AC  not  supplemental  to  FD. 
Make  DE  coincide  with  AB  :  then  EF  will 
lie  along  BC,  and  FD  must  coincide  with 
AC;  else  would  it  make  a  a  A GC  v^'xth. 
Fig  184.-  exterior  4.  AGB  =  interior   opposite  4  C, 

and  .'.  with  median  a  quadrant,  and  .*.  with  ^6^  supplemental  to 
AG,  that  is,  with  AC  supplemental  to  FD. 

404.  Theorem.  Two  spherical  triangles  of 
the    same   sense,  having  in    each    one,  and 
\  only  one,  right  angle,  equal  hypothenuses, 
and  another  side   or    angle  equal,  are  con- 
gruent. 

Proof.    If  4C=4H=x't4,  and  c  =-- h, 
Pig.  i8s.  ^nd  ^  =/,  then  if  AC  >  g,  make  CD  —  g; 

.\  BD  =  h  =  c,   and    the   bisector   of    4DBA    is  X  to  CD  A, 
,\  Bis  pole  to  CDA,  .'.  4  A  is  also  r't. 


SPHERICAL    TRIANGLES. 


77 


Fig.  i86. 


If  t  C  =  t  H  —  x\  t,  and  c  =  //,  and 
t  A  =4-  F,  then  if  ^  ABC  >  ^  G,  make 
4.  ABD  =  4.  G,  .'.  4-  BDA  =  ^  H  ^^  C  = 
r't  ^,  .:  B  is  pole  to  CD  A. 

405.  Theorem.  Of  sects  joining  two  sym- 
metrical points  to  a  third,  that  cutting  the 
axis  is  the  greater. 

Proof.  BA=BC-^  CA  =  BC  +  CA'>BA'. 

406.  Theorem.  If  two  spherical  triangles  have  two  sides  of 
the  one  equal  to  two  sides  of  the  other,  but  the  included  angles 
unequal,  then  that  third  side  is  the  greater  which  is  opposite 
the  greater  angle. 

Proof.  Slide  the  triangles  in  the  sphere  until  a  pair  of  equal 
sides  coincide  and  the  other  pair  of  equal 
sides  have  a  common  end  point.  Bisect  the 
angle  made  by  these  equal  sides.  This  axis 
cuts  the  third  side,  which  is  opposite  the 
greater  angle. 

407.  Inverse.  If  two  triangles  have  two 
sides  of   the   one  equal  to  two  sides  of  the 
other,  but  the  third  sides  unequal,  then  of 
the  angles  opposite  these  third  sides  that  is  the  greater  which 
is  opposite  the  greater  third  side. 

408.  Theorem.  The  g-line  through  the 
poles  of  two  g-lines  is  the  polar  of  their 
intersection  points. 

Proof.  If  A  and  B  are  poles  of  the  g-lines 
a  and  b,  which  intersect  in  P,  then  AP  and 
BP  diXQ  quadrants  ;    .•.  AB  is  the  polar  of  P. 

409.  Corollary  I.  The  g-line  through 
the  poles  of  two  g-lines  cuts  both  at  right 
angles. 

410.  Corollary  II.  If  three  g-lines  are  concurrent,  their  poles 
are  collinear. 

411.  Of  the  sides  of  a  spherical  angle,  we   may  call  those 


Fig.  187. 


Fig.  188. 


78 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


poles  positive  from  which   in  the  figure  these  sides  would  be 
described  from  the  vertex  by  a  quadrant  rotating  positively. 

412.  Theorem.   The  sect  which   an  angle 
intercepts  on  the  polar  of  its  vertex  equals  the 

^sect  between  the  positive  poles  of  its  sides. 

Proof.  Slide  the  quadrant  BF  along  the 
polar  of  A  until  B  comes  to  C.  The  -|-  pole 
F  oi.AB  will  then  coincide  with  the  -f-  pole  G 
oiAC. 

413.  The  sect  joining  any  point  to  one  pole 
of  a  g-line  is  less  than  a  quadrant  if  the  two 
points  are  in  the  same  one  of  that  g-line's 
hemispheres ;  greater  than  a  quadrant  if  they 
are  in  different  hemispheres. 

By  a  pole's  hemisphere  we  mean  that  one 
of  its  g-line's  hemispheres  in  which  the  pole  is. 

414.  Of  a  given  spherical  triangle  ABC, 
the  polar  is  a  new  triangle  A'B' C\  where 
A'  is  that  pole  of  56^  which  has  A  in  its 
hemisphere,  and  B'  that  pole  oi  AC  which 
has  B  in  its  hemisphere,  and  C  that  pole 
of  AB  which  has  C  in  its  hemisphere. 

415.  Theorem  If  of  two  spherical  tri- 
angles the  second  is  the  polar  of  the  first, 
then  the  first  is  the  polar  of  the  second. 

Hypothesis.  Let  ABC  be  the  polar  of 
A'B'C. 

Conclusion.  Then  A'B'C  is  the  polar 
of  ABC. 

Proof.  Join  A'B  and  A'C.     Since  B 
is  pole  of  A' C,  therefore  BA'  is  a  quad- 
rant ;  and  since  C  is  pole  of  A  'B',  therefore 
CA'  is  a  quadrant;  .*.  A'  is  pole  of  BC. 
In  like  manner,  B'  is  pole  of  AC,  and 
F.G.  X9».  C  of  AB. 


Fig.  191. 


SPHERICAL    TRIANGLES. 


79 


Moreover,  since  A  \\a.s  A'  in  its  hemisphere,  ,-.  the  sect  ^^' 
is  less  than  a  quadrant,  .*.  A'  has  A  in  its  hemisphere. 

416.  Theorem.  In  a  pair  of  polar  triangles,  any  angle  of 
either  intercepts,  on  the  side  of  the  other 
which  lies  opposite  to  it,  a  sect  which  is 
the  supplement  of  that  side. 

Proof.  Let  ABC  and  A'B'C  be  two 
polar  triangles. 

Produce  ^'^'  and  A' C  to  meet  BC  at 
D  and  E,  respectively.  Since  B  is  the 
pole  of  A'C,  therefore  BE  is  a  quadrant ; 
and  since  C  is  the  pole  of  A'B',  therefore  Fig.  193. 

CD  is  a  quadrant;  therefore  BE  -\-  CD  =^  hdM-g-Wwe-,  but 
BE -^  CD  =z  BC -\-  DE,  Therefore  DE,  the  sect  of  BC  which 
A'  intercepts,  is  the  supplement  of  BC, 

417.  Theorem.  Two  spherical  triangles  of  the  same  sense, 
having  three  angles  of  the  one  equal  respectively  to  three 
angles  of  the  other,  are  congruent. 

Proof.  Since  the  given  triangles  are  respectively  equiangu- 
lar their  polars  are  respectively  equilateral. 

For  equal  angles  at  the  poles  of  g-lines  intercept  equal 
sects  on  those  lines ;  and  these  equal  sects  are  the  supplements 
of  corresponding  sides.  Hence  these  polars,  having  three  sides 
equal,  are  respectively  equiangular,  and  therefore  the  original 
triangles  are  respectively  equilateral. 

418.  Of  a  convex  spherical  polygon  ABCD 
is    a    new   spherical   polygon    A'B'C'D'  ..., 
where  A'  is  that  pole  of  BC  which  has  A  in  its 
hemisphere,  etc. 

419.  Theorem.  The  polar  of  a  cenquad  is  a 
concentric  cenquad. 

Proof.  The  g-line  HK  through  the  sym- 
center  O  and  _L  to  ^^  is  also  _L  to  CD ;  and 
OH  =:  OK  are  the  complements  of  the  sects 
from  O  to  poles  D'  and  B'  of  the  sides  AB  and  CD. 


the  polar 


Fig.  194. 


8o 


ELEMENTARY  SYNTHETIC  GEOMETRY:. 


Hence  O  is  symcenter  for  B'  and  D'. 

In  the  same  way  prove  O  symcenter  for  A'  and  C 

420.  Theorem.  The  opposite  sides  of  a  cenquad  intersect 
on  the  polar  of  its  symcenter. 

Proof.   O  is  symcenter  for  F  and  F'. 

421.  Theorem.  Any  two  consecutive  vertices  of  a  cenquad' 
p^. ~-^c     and  the  opposites  of  the  other  two  are  con- 
cyclic. 

Proof.  The   perpendiculars,  to   the   g-Iine 
through  O  and  bisecting  BC  and  DA,  from  A 
Fig.  195.  and  B  in  one  of  its  hemispheres  and  C  and  D 

in  the  other,  are  equal.  So  also  the  perpendiculars  from  their 
opposites  D'  and  C  in  the  first  hemisphere  and  A'  and  B'  in  the 
second. 

So  A,  B,  C't  D'  and  A\  B\  C,  D  are  on  equal  circles  with 
opposite  q-poles. 

Such  circles  are  called  parallels  ;  the  co-polar  g-line,  equator. 

422.  The  perpendiculars  erected  at  the  mid  points  of  the 
sides  of  a  spherical  triangle  are  concurrent  in  its  circumcenter.^ 

423.  Theorem.  The  g-line  bisecting  two  sides  of  a  triangle 

intersects  the   third    side   at   a 
quadrant  from  its  mid  point. 

Proof.  AL,  BM,  CN  are  X 
to  the  g-Hne  through  A\  B\  the 
mid  points  of  two  sides  BC,  CA, 
D'  and  meeting  the  third  side  pro- 
duced at  n  and  B'.  .-.  ALB' 
=  CNB'  [having  the  right  angle, 
hypothenuse,  and  one  oblique 
Q  angle  equal],  :.AL  =  CN. 

Fig.  ,96.  Similarly,  BM  =  CN. 

.'.  ALD  =  BMD'  [having  two  angles  and  an  opposite  side- 
equal,  and  the  other  pair  of  opposite  sides  not  supplemental]. 
.-.  AD-  BD',  .-.  DC  a  quadrant. 


SPHERICAL    TRIANGLES.  8 1 

424.  Corollary  I.  The  altitudes  of  a  spherical  triangle  are 
concurrent  in  its  orthocenter. 

For,  regarding  A'B'C  as  the  triangle,  the  perpendicular  to 
DC  at  C  is  the  polar  of  D,  and  .*.  J.  to  A'B'. 

Similarly,  the  perpendicular  to  BA'  at  A'  is  _L  to  B'C,  etc. 

So  the  three  altitudes  of  A'B'C  are  concurrent  in  the  cir- 
cumcenter  of  ABC. 

425.  Cor,  II.  The  vertices  of  spherical  triangles  of  the  same 
angle  sum  on  the  same  base  are  on  a  circle  co-polar  with  the 
g-line  bisecting  their  sides. 

For  AO  =  BO,  ^  OAB  =  4.  OBA,  ^  LAB  =  ^  MBA 
=  i[A  +  B-{-Cl 

Hence  AOB  is  fixed,  and  .-.  (9C [supplemental  to  OA]. 

426.  Theorem.  The  g-lines  through  the  corresponding  ver- 
tices of  a  triangle  and  its  polar  are  con- 
current in  the  common  orthocenter  of  the 
two  triangles. 

Proof.  For  A  A'  is  _L  to  BC  and  B'C,  b'< 
since  it  passes  through  their  poles. 

427.  Theorem.  The  sides  of  a  triangle  ^'*^-  's?- 
intersect  the  corresponding  sides  of  its  polar  on  the  polar  of 
their  orthocenter. 

Proof.  For  AA'  is  the  polar  of  the  intersection  points  of 
BC  and  B'C  ;  similarly,  BB'  is  the  polar  of  the  intersection 
points  of  CA  and  CA',  etc. 

Sects  from  the  orthocenter  to  these  intersection  points  are 
all  quadrants.  a' 

428.  Theorem.  A  triangle's  in-center  is  also 
its  polar's  circumcenter ;  and  R  is  complemental 
to  r. 

Proof.  ID  _L  to  BC  contains  A'.     .'.  /A'  is  b" 
the  complement  of  r.     So  is  IB'  and  IC.  ^"'-  '^- 


82 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


■  I 


EXERCISES   ON   BOOK   II. 


I.  Explemental   -^s  at   the  q-poles  of  =  Os  intercept  explemental 
arcs. 


Fig   a. 


Fig.  B. 


2.  Explemental  arcs  of  equal  circles  have  equal  spherical  chords. 

3.  As  a  spherical  chord  increases  its  major 
arc  decreases. 

4.  If  Os  pass  through  2  given  p'ts  their  cen- 
ters all  lie  on  the  r't  bi'  of  the  join  of  the  2  p'ts. 

5.  If  2  Os  touch  internally,  a  1  to  the  di- 
ameter through  the  p't  of  contact  has  equal 
pieces  between  the  20  s. 

6.  The  g-lines  on  which  ±s  from  a  fixed  p't 
are  equal  envelop  a  O  with  this  p't  for  center. 

7.  The  centers  of  Os  touching  two  given 
g-lines  all  lie  on  the  bisectors  of  the  2(^s  made 
by  these  g-lines. 

8.  The  centers  of  Os  touching  3  given  g-lines 
lie  on  the  bisectors  of  the  ^s  made  by  these 
g-lines. 

9.  If  a  quad'  is  cyclic,  the  r't  bi's  of  its  sides 
and  of  its  two  diagonals  are  concurrent. 

10.  ABCD  is  a  cyclic  quad' ;  AD,  ^Cmeet  in 
F.     Where   does  tan  at  F  to  circum-o    CDF 

^G-  D'  meet  AB  ? 


EXERCISES  ON  BOOK  II. 


83 


II,  One  convex   polygon  wholly  contained  within  another  has  the 
lesser  perimeter. 


Fig.  E. 


Fig.  F. 


12.  The  perimeter  of  any  a  is  less  than  a  g-line. 

The  perimeter  of  any  convex  spherical  polygon  is  less  than  a  g-line. 

13.  If  2  Os  touch,  and  through  the  p't  of  contact  a  g-line  be  drawn 
to  cut  the  Os  again,  where  will  the  tangents  at  these  crosses  meet.? 

14.  If  2  Os  touch,  and  through  the  p't  of  contact  2  g-lines  be  drawn 
cutting  the  Os  again,  where  will  the  joins  of  these  crosses  meet.? 

15.  If  the  common  chord  of  2  intersecting  ©s  be  produced  to  any 
p't,  the  tangents  to  the  2  os  from  this  p't  are  =;  and  inversely. 

16.  If  the  common  chord  of  2  intersecting  os  be  produced  to  cut  a 
common  tangent,  it  bisects  it. 

17.  The  3  common  chords  of  3  Os  which  intersect  each  other  are 
concurrent. 

18.  How  do  the  in-,  circum-,  and  ex-radii  of  a  regular  a^  compare  in 
size? 

19.  If  a  quad'  can  have  a  circle  inscribed  in  it,  the  sums  of  the  oppo. 
site  sides  are  equal. 

20.  If  two  equal  Os  intersect,  each  contains  the  orthocenters  of  As 
inscribed  in  the  other  on  the  common  chord  as  base. 

21.  Three  equal  Os  intersect  at  a  p't  H,  their  other  points  of  inter- 
section being  A,  B,  C.  Show  that  H  is  orthocenter  of  a  ABC;  and 
that  the  triangle  formed  by  joining  the  centers  of  the  circles  is  ^  to 
A  ABC. 

22.  The  feet  of  Is  from  A  ol  A  ABC  on  the  external  and  internal 
bi's  of  ^s  B  and  C  are  co-st'  with  the  mid  p'ts  of  6  and  c. 

Does  this  hold  for  the  sphere  ? 

23.  If  two  opposite  sides  of  a  quad'  are  =,  they  make  =  ^s  with 
the  median  of  the  other  sides.  Prove  for  the  plane,  then  extend  to  the 
sphere. 


84  ELEMENTARY  SYNTHETIC  GEOMETRY. 

24.  (Bordage.)  The  ceniroids  of  the  4  As  determined  by  4  concyclic 
p'ts  are  concyclic. 

25.  The  orthocenters  of  the  4  As  determined  by  4  concyclic  p'ts 
A,  B,  C,  D  are  the  vertices  of  a  quad'  ^  to  ABCD.  The  in-centers  are 
vertices  of  an  equiangular  quad'. 

26.  CBrahmegupta.)  If  the  diagonals  of  a  cyclic  quad'  are  1,  the  J- 
from  their  cross  on  one  side  bisects  the  opposite  side. 

27.  If  the  diagonals  of  a  cyclic  quad'  are  J-,  the  feet  of  the  Is  from 
their  cross  on  the  sides  and  the  mid  p'ts  of  the  sides  are  concyclic. 

28.  It  tangents  be  drawn  at  the  ends  of  any  two  diameters,  what  sort 
of  a  quad'  is  circumscribed  } 

29.  In  any  equiangular  polygon  inscribed  in  a  ©,  each  side  is  equal 
to  the  next  but  one  to  it. 

Hence,  if  an  equiangular  polygon  inscribed  in  a  G  have  an  odd 
number  of  sides  it  must  be  equilateral. 

Any  equilateral  polygon  inscribed  in  a  ©is  equiangular. 

30.  In  any  equilateral  polygon  circumscribed  about  a  O,  each  4^ 
is  =  to  the  next  but  one  to  it. 

Hence,  if  an  equilateral  polygon  circumscribed  about  a  circle  have 
an  odd  number  of  sides,  it  must  be  equiangular. 

Any  equiangular  polygon  described  about  a  O  is  equilateral. 

31.  The  circle  through  any  3  vertices  of  a  regular  polj'gon  contains 
the  remaining  vertices. 

32.  If  one  of  2  equal  chords  of  a  0  bisects  the  other,  then  each 
bisects  the  other. 

33.  Given  2  symcentral  g-lines  and  their  symcenter.  Find  the  g-line 
symcentral  to  a  third  given  g-line  with  respect  to  this  symcenter. 

34.  All  =A'  on  the  same  side  of  the  same  base  have  their  sides 
bisected  by  the  same  g-line. 


BOOK    III. 

EQUIVALENCE. 

df2C).  Magnitudes  are  equivalent  which  can  be  cut  into  parts 
congruent  in  pairs. 

430.  Problem.  To  describe  a  rectangle,  given  two  consecu- 
tive sides. 

Construction.  Draw  a  straight,  erect  to  it  a  perpendicular. 
From  the  vertex  of  the  right  angle  lay  off  one  given  sect  on 
the  straight,  the  other  on  the  perpendicular.  Through  their 
second  end  points  draw  parallels,  one  to  the  straight,  one  to 
the  perpendicular. 

431.  Corollary,  A  rectangle  is  completely  deterrhined  by 
two  consecutive  sides  ;  so  if  two  sects,  a  and  b, 
are  given,  we  may  speak  of  the  rectangle  of  a 
and  b,  or  we  may  call    it  the    rectangle    ab. 

Thus,  when  a  and  b  are  actual  sects,  we  mean  ^ 

by  ab  a  definite  plane  figure  with  four  right  Fig.  199. 

angles,  four  sides,  and  an  enclosed  surface. 

432.  The  sum  of  two  polygons  is  any  polygon  equivalent  to 
them. 

433.  Theorem.  In  any  right-angled  triangle,  the  square  on 
the  hypothenuse  is  equivalent  to  the  sum  of  g' 
the  squares  on  the  other  two  sides.                            y^  \ 

Hypothesis,    a  ABC,  r't  angled  at  B. 

Conclusion.  Square  on  AB-\-sq^  on  BC  = 
sq'  on  AC. 

Proof.    By  430,  on  hypothenuse   AC,   on  ^ 
the  side  toward  the  A  ABC  describe  the  sq' 
ADFC. 

85 


86  ELEMENTARY  SYNTHETIC  GEOMETRY, 

On  the  greater  of  the  other  two  sides,  as  BC,  lay  off  CG  = 
AB.  Join  FG.  Then,  by  construction,  CA  =  FC,  and  AB  = 
CG,  and  ^  CAB  =  4.  FCG,  since  each  is  the  complement  of 
ACB;  .:  a  ABC  ^  a  CGF. 

Rotate  the  a  ABC  about  A  through  a  minus  r't  ^  ;  this 
brings^  to  B'.  Likewise  rotate  CGF  about  F  through  a  + 
r't  ^  ;  this  brings  G  to  G\  The  sum  of  the  angles  a.t  I)  =  st' 
^. 

.*.  G'D  and  DB'  are  in  one  straight. 

Produce  GB  to  meet  this  straight  at  H;  then  BC  =  GF  = 
FG';  and  r't  ^  6^  =  ^  GFG'  =  4  FGH\  .:  GFG'H  equals, 
square  on  BC 

Again,  BA  =  AB' ,  and  r't  ^  B'  =  ^  B'AB  =  ^  ABH\ 
.'.  ABHB  is  the  sq'  on  AB. 

.-.  sq'  o{AC=  sq'  of  AB  +  sq'  of  BC. 

434.  An  altitude  of  a  parallelogram  is  a  perpendicular  from 
a  point  in  one  side  to  the  straight  of  the  opposite  side,  which 
is  then  called  the  base. 

435.  Theorem.  A  parallelogram  is  equivalent  to  the  rect- 

G         D  F  c  angle  of  either  altitude  and  its  base. 

I -J — -I f'       ° 

j        /     I         /  Proof.  If  CD,  the  side  of  the  |g'm  opposite 

j      /       I      /  the  base  AB,  contains  F,  a  vertex  of  the  rect- 

i  /         j  /  angle,  then  ABFD  ^  ABFD,  and   A   BCF  ^ 

1/ /  A  ADG. 

A  B 

Fig.  201,  If  the  sides  AD,  BF  intersect  in  H,  then, 

G F_    D c  by  continued  bisection,  cut  BF  into  equal 

I  /  /   parts  each  less  than  BH.     Through  these 

y(  /        points  draw  straights  \   to   the  base,  so 

~/\  ~~~y  dividing  the  rectangle  into  congruent  rect- 

./- — v-^  angles,  each  as  above,  equivalent  to  the 

^  corresponding  parallelogram. 

Fig.  202-  436.  Corollary.  All  parallelograms  hav- 

ing equal  altitudes  and  equal  bases  are  equivalent. 


EQUIVALENCE. 


87 


/\ 


\ 


^F 


437;  Theorem.  A  triangle  is  equivalent 
to  the  1  bctangle  of  its  base  and  half-alti- 
tude. 

Procf .  Join  the  mid  points  D,  M  of  the  Fig.  203. 

sides  CB,  BA  of   A  ABC,   and  produce  MD'  =  MD.     Then 
aAD'M^  aBDM. 

438.  Corollary.  Triangles  of  equal  bases  and  altitudes  are 
equivalent. 

439.  Theorem.  A  trapezoid  is  equivalent 
to  a  triangle  of  equal  altitude,  whose  base 
is  the  sum  of  the  parallel  sides. 

Proof.  Join  AC,  BD.     To  DA  produced  ^ ^^ 

dr2iw  BF\to  CA.  F.G.204. 

A  BCD  =  A  BCA  ^  A  AFB. 

440.  Theorem.  The  sect  cut  out,  on  a 
parallel  to  the  base  of  a  triangle  by  the  sides, 
is  bisected  by  the  corresponding  median. 

Proof.  Let  AI  be  the  mid  point  of  PQ  \  to 
AB.  t  PMC  =  t  QMC\  also  trapezoid 
AC'MP=  trapezoid  C'BQM;  .'.  AC  MCA  = 
CBCMC. 

Were  M  not  in  CC,  but  on  Q's  sivle,  then  AC  MCA  >  A 
ACC>  CBCMC. 

441.  Theorem.  Sects  joining  intersections  of  the  sides  of  a 
parallelogram  with  straights  drawn  parallel  d    p    rg        c 
to  the  sides  through  a  point  on  one  diagonal, 
if  they  cut  that  diagonal,  are  parallel  to  the 
other. 

Proof.  Through  O  draw  QR  \\  to  BC, 
cutting  HK  in  5.  Since  DR  =  RC,  .:  MS 
=  S£,  and  HM  =  EK=FB',  .:  HMBF  is  a  ||g'm. 

Again,  since  MK  =  HE  =  DG,  .-.  MKGD  is  a  ||g'm. 

442.  Corollary.  If  through   any  point  on  a  diagonal  of  a 
parallelogram  straights  be  drawn  parallel  to  the  sides,  the  two 


88  ELEMENTARY  SYNTHETIC  GEOMETRY. 

parallelograms,  one    on    each    side    of   this   diagonal,  will    be 
equivalent. 

For  through  E  drawing  NP  \  to  BD,  we  get  a  FBL  ^  A 
HMD,  and  a  ENK  ^  a  PEG,  and  Ig'm  BNEL  =  ||g'm 
DMEP. 

443.  Theorem.  Any  angle  made  with  a  side  of  a  spherical 
triangle  by  joining  its  extremity  to  the 
circumcenter,  equals  half  the  angle-sum 
less  the  opposite  angle  of  the  triangle. 

Proof.     For    ^.A-^^B^tC—  2 
4-  OCA  +2  4.  OCB  ±  2  ^  OAB 

.'.  ^  OCA  =\ltA^^B-\-tC\- 
[^  OCB  ±  ^  OAB]  =  ^[^  A  -^  4  B 
F,o..o,.  +^C]-^B. 

444.  Corollary.  Symmetrical  spherical  triangles  are  equiva- 
lent. 

For  the  three  pairs  of  isosceles  triangles  formed  by  joining 
the  vertices  to  the  circumcenters,  having  respectively  a  side 
and  two  adjacent  angles  equal,  are  congruent. 

445.  Theorem.  When  three  g-lines  mutually  intersect,  the 
two  triangle^  on  opposite  sides  of  any  vertex  are  together 
equivalent  to  the  lune  with  that  vertical  angle. 

Proof.    A   ABC  -{-  A   ADF=  lune 
ABHCA. 

For  DF  —  BC,  having  the  common 
Id  supplement    CD ;    and  FA  =  CH,  hav- 
ing the  common  supplement  HF\  and 
AD  =  BH,  having  the  common  supple- 
^  ment  ND;  .:    2  ADF  =  2  BCH ,  .: 

Fig.  208  a.  A    ABC  +   A  ADF  =  2  ABC  4-   A 

BCH  =  \\xnQ  ABHCA. 

446.  The  spherical  excess,  e,  of  a  spherical  triangle  is  the  ex- 
cess of  the  sum  of  its  angles  over  a  straight  angle.     In  general. 


:  EQUIVALENCE.  89 

the  spherical  excess  of  a  spherical  polygon  is  the  excess  of  the 
sum  of  its  angles  over  as  many  straight  angles  as  it  has  sides, 
less  two. 

447.  Theorem.   A  spherical  triangle  is  equivalent  to  a  lune 
whose  angle  is  half  the  triangle's  spher- 
ical excess. 

Proof.   Produce  the  sides  of  the  ^ 
ABC  until  they  meet   again   two   and  b| 
two  at  D,  F,  and  H.     The  ^  ABC  now 
forms  part  of  three  lunes,  whose  angles 
are  A,  B,  and  C,  respectively. 

But,  by  436,   lune   with    ^  A  =  £\  fTg.  2o87. 

ABC-\-2ADF. 

Therefore  the  lunes  whose  angles  are  A,  B,  and  C  are  to- 
gether equal  to  a  hemisphere  plus  twice  a  ABC.  But  a 
hemisphere  is  a  lune  whose  angle  is  a  straight  angle;  .*.  2 a 
ABC  =  lune  whose  ^  is  [A  -\-  B  -\-  C  —  st.  ^  ]  =  lune  whose 
^  is  e. 

448.  Corollary  I.  The  sum  of  the  ^s  of  a  a  is  >  a  st'  ^ 
^nd  <  3  st'  ^  s. 

449.  Cor.  II.   Every  :^  of  a  a  is  >  ^e. 

450.  Cor.  III.  A  spherical  polygon  is  equivalent  to  a  lune 
whose  angle  is  half  the  polygon's  spherical  excess. 

451.  Cor.  IV.  To  construct  a  lune  equivalent  to  any  spheri- 
cal polygon,  add  its  angles,  subtract  [w  —  2]  st'  ^s,  halve  the 
remainder,  and  produce  the  arms  of  a  half  until  they  meet 
again. 


90  ELEMENTARY  SYNTHETIC  GEOMETRY. 


EXERCISES  ON  BOOK   III. 

I.  The  joins  of  the  centroid  and  vertices  of  a 
triangle  trisect  it. 

Proof.    A  ABM  =  A  MBC, 

A  AGM  =  A  MGC\ 

.-.  A  ABO  =  A  GBC. 

I.  Make  a  ||g'm  triple  a  given  ||g'm. 

3.  Make  a  A  triple  a  given  A. 

4.  Make  a  symtra  triple  a  given  symtra. 

5.  Trisect  a  given  symtra. 

6.  {a  +  by  t:  a'  +  zab  +  ^». 

7.  {a  —  by  =  a'  -  2ab  +  P. 

8.  {a  +  b){a  —  b)  =  a"  -  b\ 

10.  On  each  side  of  a  quad'  describe  a  sq'  outwardly.  Of  the  four  as 
made  by  joining  their  neighboring  corners,  two  opposite  equal  the  other 
two  and  equal  the  quad. 

II.  (Pappus.)  Describe  on  two  sides  AB,  AC  of  a  A  any  ||g'ms 
(both  outwardly  or  both  inwardly).  Designate  the  cross  of  the  sides 
opposite  b  and  c  by  F.  On  the  st'  FA  from  a  cut  off  A'//  =  AF.  Con- 
struct a  II  g'm  on  a  whose  opposite  side  goes  through  //.  It  equals  the 
sum  of  the  other  two. 

12.  If  from  an  ^  a  we  cut  two  =  as,  one  •!• ,  the  sq'  of  one  of  the  = 
sides  of  the  -I-  A  equals  the  rectangle  of  the  sides  of  the  other  a  on  the 
arms  of  the  ^  a. 

13.  Transform  a  given  A  into  an  =  -j-  a. 

14.  Transform  a  given  a  into  an  =  regular  a. 

15.  If  a  vertex  of  a  A  moves  on  a  ±  to  the  opposite  side,  the  differ- 
ence of  the  squares  of  the  other  sides  is  constant. 

16.  The  ^  bisectors  of  a  rectangle  make  a  sq',  which  is  half  the  sq' 
on  the  difference  of  the  sides  of  the  rectangle. 

17.  The  bisectors  of  the  exterior  2(1  s  of  a  rectangle  make  a  sq'  which 
is  half  the  square  on  the  sum  of  the  sides  of  the  rectangle. 

18.  The  sura  of  the  squares  made  by  the  bisectors  of  the  interior  and 


EXERCISES  ON  BOOK  III.  9r 

exterior^  ^of  a  rectangle  equals  the  square  on  its  diagonal ;  their  differ- 
ence is  double  the  rectangle, 

19.  If  on  the  hypothenuse  we  lay  off  from  each  end  its  consecutive 
side,  the  sq'  of  the  mid  sect  is  double  the  rectangle  of  the  others. 

20.  If  in  A  ABC,  the  foot  of  altitude  from  A  be  D,  from  C  be  ^,  then 
rectangles  BD  .  a  =  BF .  c. 

(Hint.  From  4-  B  are  two  r't  as  cut  off.  Turn  one  about  the  bi- 
sector of  4(!  B.) 

21.  In  a  trapezoid,  the  sum  of  the  squares  on  the  diagonals  equals 
the  sum  of  the  squares  on  the  non-J  sides  plus  twice  the  rectangle  of  the 
H  sides. 


BOOK    IV. 

PROPORTION. 

452.  A  greater  magnitude  is  said  to  be  a  multiple  of  a  lesser 
magnitude  when  the  greater  is  the  sum  of  a  number  of  parts 
each  equal  to  the  lesser;  that  is,  when  the  greater  contains  the 
lesser  an  exact  number  of  times.  The  lesser  is  then  called  a 
submultiple  of  the  greater. 

453.  Any  multiple  of  any  submultiple  of  a  magnitude  is 
called  d,  fraction  of  that  magnitude. 

454.  Two  magnitudes  of  which  neither  is  a  fraction  of  the 
other  are  called  incominensiirable ;  for  example  i  and   4^2. 

455.  That  definite  mmterical  relation  of  any  magnitude  to 
any  magnitude  of  the  same  kind,  in  virtue  of  which  the  former 
is  either  a  fraction  of  the  latter  or  is  greater  than  one  and  less 
than  the  other  of  two  fractions  of  the  latter  differing  by  less 
than  any  given  fraction  however  small,  is  called  the  ratio  of 
the  former  to  the  latter. 

456.  If  the  first  of  two  magnitudes  is  a  fraction  of  the 
second,  the  ratio  of  the  former  to  the  latter  is  expressed  by  the 
numerical  fraction  whose  denominator  is  the  number  indicating 
the  submultiple  of  the  second,  and  whose  numerator  is  the 
number  indicating  the  multiple  of  that  submultiple. 

Thus  the  ratio  of  a  foot  to  8  inches  is  3/2. 

457.  The  ratio  of  the  first  of  two  magnitudes  to  the  second 
is  said  to  be  greater  than  a  numerical  fraction  expressing  the 
ratio,  to  the  second,  of  any  magnitude  less  than  the  first. 

458.  Two  ratios  are  equal  if  no  numerical  fraction  is  greater 
than  one  and  less  than  the  other. 

459.  When  the  ratio  of  two  magnitudes  A  and  B,  which 

92 


PROPORTION.  95. 

may  be  written  A/B,  equals  that  of  the  other  two  a  aiid  b,  the. 
four  are  said  to  form  a  proportion ;  which  may  be  written 
A/B  =  a/b. 

460.  Theorem.  If  to  every  one  of  a  series  of  magnitudes 
A,  B,  C,  .  .  .  there  corresponds  one  of  a  second  series. «,  b,  c,  .  .  . 
in  such  manner  that, 

1.  If  the  magnitudes  A  and  B  are  equal,  so  are  also  the. 
corresponding  magnitudes  a  and  b  ;  and, 

II.  The  sum  vS"  of  two  magnitudes  A  and  B  corresponds  to 
the  sum  s  of  the  corresponding  magnitudes  a  and  b', 

Then  two  magnitudes  of  the  first  series  have  the  same  ratio, 
as  the  corresponding  magnitudes  of  the  second  series. 

Proof.  I.  If  B  corresponds  to  b,  and  7t  is  any  integer,  then 
tiB  corresponds  to  7ib  ;  for  the  sum  of  ti  equal  parts  B  must  [by 
II]  correspond  to  the  sum  of  11  equal  parts  b. 

2.  Also  the  wth  part  of  B  corresponds  to  the  «th  part  of  b; 
for  a  magnitude  which,  taken  11  times  gives  B  must  correspond 
to  that  which  taken  71  times  gives  b. 

First  Case.  When  ^  is  a  fraction  of  B. 
Then  A  =  {7i/d)B  =  7t .  {B/d). 

Now  [by  2]   the  magnitude  B/d  corresponds  to  b/d,  an(\ 
[by  i]  the  magnitude  7i.{B/d)  corresponds  to  7i{b/d). 
Consequently  a  =  71.  [b/d)  =  {7i/d)b. 
Second  Case.  When  A  is  no  fraction  of  B. 
Then  U  A>  {7i/d)B,   .:  A/B  >  {n/d)B/B,   .'.  A/B  >  7t/d.. 

71  71 

But  since  A  >  -B,  .-.  [by  II]  a^  —b,  .-.  a/b  >  7t/d. 
d  d 

461.  Corollary  I.  If  parallels  cut  two  straights,  the  inter-, 
cepts  on  one  have  the  same  ratio  as  the  cor- 
responding intercepts  on  the  other. 

For  to  sects  a,b,c,...  on  one,  the  parallels 
give  corresponding  sects  a',  b',  c' ,  .  .  .  on  the 
other,  such  that  \{  a  =^  b,  then  a'  =^  b'  \  and  to 
the  sum  a-\-b  corresponds  the  sum  a'  -\-  b' ,  etc.  '        jne.  309. 


•94  ELEMENTARY  SYNTHETIC  GEOMETRY, 

462.  Cor.  II.   Parallelograms  with  an  angle  and  a  side  in  one 

equals  to  an  angle  and  side  in 
the  other  have  the  same  ratio  as 
their  other  sides. 

For  this  other  side  and  the  ||g'm 
are  then  corresponding  magni- 
tudes, such  that  if  of  sides  a,  b,  c,  .  .  .  and  ||g'ms  A,  B,  C,  .  .  . 
/I  z=  d,  .'.  A  =  B,  also  \.o  a  -\-  b  corresponds  A  ■\-  B. 

463.  Cor.  III.  In  the  same  circle  or  in  equal  circles,  angles 
at  the  center  have  the  same  ratio  as  their  arcs. 

For  these  angles  A,  B,  C,  .  .  .  and  arcs  a,  b,  c,  .  .  .  so  corre- 
spond that  if  A  =  B,  then  a  =  b  ;  and  to  A  -\-  B  corresponds 
^i-\-b. 

464.  Chords  are  not  proportional  to  their  arcs. 

For  if  arcs  A,  B  correspond  to  chords  a,  b,  then  arc  A  -\-  B 
does  not  correspond  to  a  chord  equal  to  a-\-  b. 


BOOK   V. 

SIMILARITY. 

465.  If  from  a  point  we  draw  rays  to  all  the  points  of  a 
given  figure,  and  take  on  each  of  these  rays  another  point, 
these  latter  points  determine  a  second  figure  which  we  may 
call  a  perspective  of  the  given  figure.  Two  figures  are  called 
perspective  when  each  point  of  one  can  be  so  paired  with  a 
point  of  the  other  that  the  joins  of  all  the  pairs  concur  in  one 
point  called  the  center  of  perspective.  Two  figures  are  called 
projective  if  they  can  be  moved  so  as  to  be  perspective. 

Two  figures  are  called  similar  when  they  can  be  so  placed 
that  on  any  straight  whatever  through  one  point  sects  from  it 
to  the  perimeters  of  the  figures  have  always  the  same  ratio. 

Figures  are  similar  which,  being  projective,  when  made  per- 
spective have  sects  from  the  center  of  perspective  to  the  pairs 
of  points  always  in  the  same  ratio. 

466.  The  sect  from  the  center  of  perspective  to  any  point 
is  called  that  Rami's  perspective  sect. 

467.  A  point  in  the  perimeter  of  one  figure  and  a  point  in 
the  perimeter  of  the  other  are  said  to  correspond  if  their 
perspective  sects  are  co-straight  when  the  figures  are  in 
perspective  position.  Should  the  perimeters  then  have  four 
points  co-straight,  of  the  two  on  one  figure,  that  whose  per- 
spective sect  is  the  lesser  corresponds  to  that  of  the  two  on 
the  other  figure,  whose  perspective  sect  is  the  lesser. 

468.  The  ratio  of  corresponding  perspective  sects  is  called 
the  ratio  of  similitude  of  similar  figures  ;  the  perspective  center, 
their  center  of  similitude. 

469.  The  center  of  perspective  is  called  internal  when 
corresponding  perspective  sects  lie  on  opposite  sides  of  it ; 
otherwise,  external. 

95 


96 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


Fig. 


Fig. 


A  symccnter  is  that  special  case  of  an  internal  center  of 
perspective  where  the  corresponding  perspective  sects  are; 
equal  in  magnitude. 

470.  Theorem.  Any  two  circles  are  similar  figures. 

Proof.  When  concentric,  their  center  is  a  center  of  perspec- 
tive, and  the  ratio  of  corresponding  perspective  sects  is  constant, 
being  the  ratio  of  the  radii. 

471.  Corollary.  The  intersection  point  of  two  straights; 
is  '^  {7  for  the  arcs  they  intercept  on  circles 
with  that  point  as  center. 

For  the  ratio  of  the  radii  gives  a  constant; 
ratio  of  similitude. 

"S  472.    The    intersection    point     of    two 

straight,  is  ~  (7  for  the  sects  they  cut  out, 
on  any  two  parallels. 

For  a  parallel  through  this  point  shows 
a  constant  ratio  of  similitude. 

473.  Theorem.  If  points  on  two 
straights  are  made  corresponding  which 
end  sects  from  their  intersection  having 
the  same  ratio,  the  straights  through  cor- 
responding points  [projection-straights] 
are  parallel. 

Proof.  By  hypothesis  CA/CA'  =  CB/CB'\  .-.  CA/CB  =-. 
CA'/CB'.  But  the  parallel  to  A  A'  drawn  through  B  gives- 
CA/CB  =  CA'/CB". 

.-.  B"  coincides  with  B' . 

A'B.'  474.  Theorem.  Similar  sects 

have  the  ratio  of  similitude. 
Proof.    When    in    perspec 
b'  tive  position  since   CA/CB  = 
Fig.  214.  Fig.  215.     O^'/C^',.'. thesectsareparallel. 

Now  slide  CBB'  until  B'  comes  on  A'  and  BB'  contains 
A.  Thus  A'  becomes  ~  C  for  the  parallels  AC  and.  BC  '„ 
,'.  B'B/A'A  =  B'C'/A'C. 


Fig.  213. 


SIMILARITY.  97 

475.  Problem.  To  three  given  sects  to  find  a  fourth  propor- 
tional. 

Construction.  On  one  arm  of  any  ^  C  cut  off  CD  ■=.  a, 
and  DF  =^  b\  on  the  other  arm  make  CH  =.  c.  Join  DH. 
Draw  FK  \  to  DH.     a/b  =  c/  {_HK\ 

476.  We  say  that  by  a     ^  ^ 
point  Pox\  the  straight  AB, 

but   not  on  the  sect  AB,  \\\ ^ 

this  sect   is  divided  extcr- 

nally;  and  y^/* and  ^/^  are       fig.  216. 

called  exter7ial  segments  of  the  sect  AB. 

If  the  point  Pis  on  the  sect  AB,  this  is  said  to  be  divided 
internally. 

477.  Problem.  To  divide  a  given  sect  AB  in  a  given  ratio, 
AS/BT. 

Construction.  On  parallels,  from  A  and  B  take  on  opposite 
sides  of  the  straight  AB  [or  the  same  side]  sects  AS  and  BT. 
Join  57;  cutting  AB  in  P.     Then  AP/PB  =  AS/BT. 

478.  When  a  sect  is  divided  internally  and  externally  into 
segments  having  the  same  ratio,  it  is  said  to  be  divided  /lar- 
tnonically. 

479.  Theorem.  If  a  sect  ^^  is  divided  harmonically  by  the 
points  Pand  Q,  the  sect  PQ  will  be  divided  harmonically  by 
the  points  A  and  B. 

Proof.  Since  AP/BP  =  AQ/BQ, 
.:BP/AP=BQ/AQ; 
.-.BP/BQ  =  AP/AQ. 

480.  The  points  A,  B,  and  P,  Q,  of  which  each  pair  divides 
harmonically  the  sect  terminated  by  the 

■'  ^  A P      B Q, 

other  pair,  are  called  four  harmonic  points 

/  •  Fig.   218. 

or  a  harmomc  range. 

481.  Problem.  Given  a  point  P  on  the  straight  AB,  to  de- 
termine the  fourth  harmonic  point. 

Construction.  Through  A  and  B  draw  parallels,  and  by  a 


98 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


straight  through  P,  cut  them  in  vS  and    T.     On  straight  BT 
take  BT  =  BT.     The  fourth  point  is  on  the  straight  ST'. 

482,  Corollary.  With  ABP  only  one  point  forms  a  har- 
monic range,  for  if  S  be  any  point  without  the  straight  in 
which  is  the  harmonic  range  APBQ,  and  if  through  B  we 
draw  BT  II  to  AS,  meeting  SPm  T,  SQ  in  T,  then  BT  =  BT . 
483.  Theorem.  If,  of  two  parallels,  those 
points  which  end  proportional  sects  be  joined, 
these  projection-straights  are  concurrent. 

Proof.  Let  the  projection-straights  A  A'  and 
BB'  meet  in  C.     Then  let  CD  cut  A' B'  in  F.     By 
hypothesis  AB/BD  =  A'B'/B'D'.  By 472  AB/BD 
=  A'B'/B'F.     Hence  F coincides  with  D'. 

484.  Theorem.  In  any  trapezoid  the  mid  points  of  the 
parallel  sides  and  the  intersection  point  of  the  non-parallel 
sides  and  the  intersection  point  of  the  diagonals  form  a  har- 
monic range. 

p  „  Proof.  P  divides  the  median  AB  in 

the  ratio  of  the  parallel  sides. 

485.  Theorem.  The  medians  of  a 
triangle  are  concurrent  in  that  trisection 
A  point  of  each  remote  from  its  vertex. 
Proof.  The  sect  joining  the  mid 
points  of  two  sides  of  a  triangle  is  |I  to  and  ^  of  the  third  side ; 
.-.  the  intersection  point  of  any  two  medians,  since  it  divides 
each  median  in  the  ratio  of  these  \s,  is  that  trisection  point  of 
each  remote  from  its  vertex. 

486.  The  intersection  point  of  its  medians  is  called  the  tri- 
angle's centroid. 

487.  Theorem.  The  bisector  of  an  exterior  angle  or  in- 
terior angle  of  a  triangle  divides  the  opposite  side  externally 
or  internally  in  the  ratio  of  the  other  two  sides  of  the 
triangle. 

Proof.  ABC  any   A,  BD  the  bisector  of  ^  at   B.     Draw 


Fig.  221. 


SIMILARITY.  99 

AF II  BD.  Then  of  the  two  angles  at  B  given  equal  by- 
hypothesis,  one  equals  the  correspond- 
ing interior  angle  at  F,  and  the  other 
the  corresponding  alternate  angle  at  A^ 
.'.  AB  =  BF  [sides  opposite  equal  ^s]. 
But  BF/BC  =  AD/ DC.  [If  parallels 
cut  two  straights,  their  intercepts  are  proportional.] 

.-.  AB/BC  =  AD/DC. 

488.  Inverse.  If  one  side  of  a  triangle  is  divided  inter- 
nally or  externally  in  the  ratio  of  the  other  sides,  the  straight 
from  the  point  of  division  to  the  opposite  vertex  bisects  the 
interior  or  exterior  angle. 

489.  Corollary.  The  bisectors  of  an  interior  and  exterior 
angle  at  one  vertex  of  a  triangle  divide  the  opposite  side  har- 
monically. 

490.  Theorem.  Two  triangles  are  similar  if  they  have  two 
angles  respectively  equal,  or  two  sides  proportional  and  the 
included  angles  equal,  or  two  sides  proportional  and  the  angles 
opposite  the  greater  equal,  or  their  three  sides  proportional. 

Proof.  Put  one  angle  upon  its  equal,  and  then  the  common 
vertex  is  ~  C 

The  A  with  three  sides  proportional  to  those  of  a  given  A 
is  ^  to  the  A  made  by  a  straight  ||  to  one  side  of  the  given  a  , 
and  cutting  off  from  a  second  side  a  sect  equal  to  the  corre- 
sponding side  of  the  other  a. 

491.  Theorem.  In  a  right  triangle  the  altitude  to  the 
hypothenuse  is  a  mean  proportional  between  the  segments  of 
the  hypothenuse,  and  each  side  is  a  mean  proportional  between 
the  hypothenuse  and  its  adjacent  segment. 

Proof.  R't  /\ABC '^  £,ACD  ^  aCBD. 

492.  Corollary.  To  find  a  mean  proportional  to  two  given 
sects,  put  a  semicircle  on  their  sum  as  diameter,  and  produce 


lOO 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


to  this  semicircle  the  perpendicular  erected  at  their  common' 
point. 

493.  Theorem.  If  four  sects  are  proportional,  the  rectangle 
contained  by  the  extremes  is  equivalent  to  the  rectangle  con- 
tained by  the  means. 

Proof.  Let  the  four  sects  a,  b,  c,  d  be  proportional. 

On  a  and  on  b  construct  rectangles  with  altitude  c.     On  c 


a 

ac 
c 

a 

nd 

d. 

ac 
a 

c 

i  k 

1       6 

■C 

Fig.  223. 

and  on  d  construct  rectangles  of  altitude  a.    Then  a/b  =  ac/bc^ 
and  c/d  =  ac/ad. 

[Rectangles  of  equal  altitudes  are   to  each  other  as  their 
bases.]     But  by  hypothesis,  a/b  =  c/d. 

.'.  ac/bc  =  ac/ad;     .'.  be  =  ad. 

494.  Theorem.  The  rectangle  of  the  segments  into  which  a 
given  point  divides  chords  of  a  given  circle  is  constant. 

p 


Fig.  224. 


Fig.  225. 


Hypothesis.  Let  chords  AB  and  CD  intersect  in  P. 
Conclusion.  Rectangle  AP.PB  =  rectangle  CP.PD. 
Proof,    i.  PAC  —  ^  PDB   [inscribed    angles  on  the  same 


SIMILARITY. 


lOI 


arc],   and    t  APC  =  ^  BPD- 
angular  triangles]. 


t.APC'^  A  BPD     [equian- 


FlG. 


.-.  AP/CP  =  PD/PB-    .-.  AP.PB  =  CP.PD. 

495,  Corollary.  Let  the  point  P  be 
without  the  circle,  and  suppose  DCP 
to  revolve  about  P  until  C  and  D  coin- 
cide: then  the  secant  DCP  becomes 
a  tangent,  and  the  rectangle  CP.PD 
becomes  the  square  on  PC.  There- 
fore, if  the  point  is  without  the  circle, 
the  rectangle  is  equivalent  to  the 
square  of  the  tangent ;    if  within,  to  the  square  on  half   the 

smallest  chord. 

496,  Theorem,  If  a  triangle  have  two  sides 
each  equal  to  c,  and  from  their  intersection  a 
sect  d  cut  the  third  side  into  segments  /"and 
^,  then  e  =  d'  -^fg. 

Proof.         ^"  =  //^  +  [K/+^)-/r; 

,.d^  =  h^-^\k{g-f)-\\        ■  -^  .: 

But  ^  =  >^'  +  [K^ +  /)]'. 

497,  Theorem.  A  point  without  a  circle,  and  its  chord  of 
•contact,   divide  harmonically   any   chord 
whose  straight  contains  the  point. 

Proof.  AP.PB  =  €'  =  PQ' -\-CQ.QD 
■^PQ^AQ.QB. 

But  AP.PB  =  AP'  +  AP.AQ  + 
AP.QB:  and  PQ'  -\-AQ.QB  =  AP" 
-\- AP.AQ -\-AQ.PQ  +  AQ.QB. 

.:AP.QB  =  AQ.PB;  .:  AP/PB  = 
AQ/QB. 


Fig.  228. 


I02  ELEMENTARY  SYNTHETIC  GEOMETRY. 

498.  Theorem.  The  rectangles  of  opposite  sides  of  a  noiv 
cyclic   quadrilateral   are  together  greater 
than  the  rectangle  of  its  diagonals. 
^        Proof.    Make  4.  BAF  —  ^  CAD,    and 
tABF=4.ACD. 

^"^^^    ^^   I  Join  FD. 

Then       aABF^^  aACD, 


Fig.  229^. 


.-.  BA/AC  =  FA/AD. 


But  this  shows  (since  14.  BAC  =  t  F^D), 


ABAC  -^  aFAD. 


From 


aABF -^  aACD, 
.'.  AB.CD  =  BF.AC. 


From 


ABAC '^  aFAD, 
.'.  BC.AD=  FD.AC. 


\<  ab.cd-\-bc.ad  =  bf.ac-\-fd.ac>bd.ac: 


499.  Corollary  (Ptolemy).  The  rectangle  of  the  diagonals. 
of  a  cyclic  quad'  equals  the  sum  of  the  rectangles  of  opposite 
sides. 

For  then  F  falls  on  BD. 


EXERCISES  ON  BOOK   V-  I03 


EXERCISES  ON   BOOK  V. 

1.  The  joins  of  the  vertex  of  one  ^  of  a  A  to  the  ends  of  that  diam- 
eter of  the  circum  O  which  is  J.  to  the  opposite  side  are  the  bisectors  of 
that  t- 

2.  If  2  AS  have  a  common  base,  they  are  as  the  segments  into  which 
the  join  of  the  vertices  is  divided  by  the  common  base. 

3.  The  3  external  bisectors  of  the  ^s  of  a  A  meet  the  sides  co- 
straightly. 

4.  Given  one  side  of  a  A,  and  the  ratio  of  the  other  sides,  find  the 
path  of  its  movable  vertex. 

5.  The  sect  ||  to  one  side  of  a  quad'  from  the  cross  of  2  diagonals  and 
bisected  by  the  opposite  side  ends  where  .-* 

6.  If  equiangular  as  have  a  common  vertex  and  second  vertices  co- 
st', so  are  the  third  vertices. 

7.  If  c  be  the  center-sect  of  the  in-  and  circum- ©s  of  a  A,  then 


R  +  c     R  —c 


=  I. 


8.  The  OS  on  the  sides  of  a  a  as  diameters  cross  on  the  sides  of  the  A . 

9.  If  a  o  be  described  on  one  of  the  X  sides  of  a  r't  A  as  diameter, 
the  tangent  at  the  p't  where  it  divides  the  hypothenuse  bisects  the  other 
±  side. 

10.  The  mid  p'ts  of  concurrent  chords  are  concyclic. 

11.  A  A',  BB',  CC  are  ||  chords  of  a  ©.     Show  A  ABC  \  A  A'B*C'. 

12.  AB  is  trisected  in  C  and  D ;  CPD  is  a  regular  A  ;  show  that  D  is 
circumcenter  of  BPC,  and  AP  the  tangent  at  P  to  the  circum- 0. 

13.  Two  AS  on  opposite  sides  of  the  same  base  have  the  ^s  opposite 
it  supplemental.  Show  that  the  join  of  their  supplemental  ^s  is  j  to 
the  join  of  their  orthocenters. 

14.  If  the  ±  projections  of  any  vertex  of  a  quad'  on  the  other  sides 
and  diagonal  of  the  quad'  are  co-straight,  so  are  the  like  projections  of 
any  other  vertex. 


BOOK   VI. 

MENSURATION. 

500.  In  practical  science,  every  quantity  is  expressed  by 
another  of  the  same  kind  preceded  by  a  number. 

From  our  knowledge  of  the  number  and  the  quantity  it 
multiplies,  we  get  knowledge  of  the  quantity  to  be  expressed. 

So  in  each  kind  of  magnitude  we  select  one  convenient 
quantity  as  a  standard  or  unit,  to  be  known  familiarly  by  us, 
and  then  to  be  used  in  expressing  every  other  magnitude  of  the 
same  kind. 

The  measurement  of  a  magnitude  consists  in  finding  its  ratio 
to  its  unit. 

501.  For  sects,  the  unit  is  the  centimeter  ['^'"•],  which  is 
the  hundredth  part  of  the  sect  between  two  marked  points  on 
a  special  bar  of  platinum  at  Paris,  when  the  bar  is  at  the  tem- 
perature of  melting  ice. 

The  length  of  any  sect  is  its  ratio  to  the  centimeter. 

502.  An  accessible  sect  may  be  approximately  measured  by 
the  direct  application  to  it  of  a  centimeter,  or  a  sum  of  centi- 
meters, such  as  the  edge  of  a  ruler  suitably  divided. 

But  because  of  incommensurability,  even  were  our  senses 
perfect,  any  direct  measurement  must  be  usually  imperfect  and 
merely  approximate. 

503.  For  the  measurement  of  surfaces  the  standard  is  the 
square  centimeter  [*"""],  the  square  on  the  linear  unit. 

504.  The  area  of  any  surface  is  its  ratio  to  this  square. 

104 


MEN  SURA  TION. 


105 


505.  Theorem.  The  area  of  a  rectangle  equals  the  product 
■of  the  length  of  its  base  by  the  length  of  its  altitude. 

Proof.  If  the  altitude  of  the  rectangle  R  is  a,  and  its  base  b, 
then  its  ratio  to  a  rectangle  of  altitude  i*^™  and  base  tF^  is  a', 
but  the  ratio  of  this  rectangle  to  the  square  centimeter  is  h ; 

.-.  R  —  ab'''^\ 

506.  Corollary.  The  area  of  a  square  is  the  second  power  of 
the  number  denoting  the  length  of  its  side. 

507.  Cor.  From  the  area  of  a  square,  to  find  the  length 
of  its  side  :  extract  the  square  root  of  its  area. 

508.  To  find  the  length  of  the  other  side,  from  the  length 
of  the  hypothenuse  and  of  one  side  of  a  right  triangle,  multiply 
the  sum  of  the  lengths  by  the  difference,  and  extract  the  square 
root. 


C  —  a- 


\c  -  a\\c -^ d"^  =  b\ 


509.  Given   the  chord  of   an  arc  and   the 
radius  of  the  circle,  to  find  the  chord  of  half  g 
the  arc. 

BC  =k'  =  Vbd'  +  DC  =  '/[py+:^». 


Fig.  229. 


But 
and 


DO  =  iOC  -  ODY  =  [r  -  ODY, 


OD  =  ^OB"  -  30-"  =  Vr'  -  -. 

4 


.-.  k'  =  \/--\-{r-  \^r^  -  -S  =  V^2r'  -  2r/; 
4  4^ 


510.  Since  the  angle  at  the  center  subtended  by  the  side  of 
a  regular  inscribed  hexagon  is  one  third  a  straight  angle,  and 


io6 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


SO  this  side  equals  the  radius,  therefore  the  length  of  a  side  of 
a  regular  dodecagon  inscribed  in  a  circle  whose  radius  is  i<='"  is 


k'  =  ^2  —  2Vi—i  — -  0.517638C9+. 

The  length  of  one  side  of  a  regular  inscribed  polygon  of  24 
sides  is 

r' = /rryi-(ii7Q38°9+)'^  0.26,05238 +. 

4 

511.  Given  the  radius  of  a  circle  and  the  side  of  a  regular 
inscribed  polygon,  to  find  the  side  of  a 
similar  circumscribed  polygon. 

Suppose  AB  the  given  side  k.    Draw 
the  tangent  at  the  mid  point  C  of  the 
arc  AB,  and  produce  it  both  ways  to 
the  points  D  and  F,  where  it  meets  the 
Fig.  230,  j.^(jii  Q^  ^LU^  QB  produced.     DF  is  the 

side  required,  t. 

t.OFC  ^  /\OBM, 
.'.  OC/OM=CF/BM=t/k. 

kr 


But 


t  = 

0M  = 


OM' 


VOB'  -  BM'  =  Vr'  -  \k\ 
kr 


V'i 


512.  Corollary.  When  r  =  i"",  the  side  of  a  regular  circum* 
scribed  hexagon 

I        _    2 

vr^~  4/3' 


/  = 


MENSURA  TION. 


107 


The  side  of  a  regular  circumscribed  dodecagon 

.  0.51763809  „  o   , 

t'  =  ^  y  ^  ^        =  .535898  +. 


Vi 


i(.  5 1 763809)' 


513.  Since  no  part  of  a  circle  can  be  congruent  to  any  sect, 
so  no  part  of  a  circle  can  be  equivalent  to  any  sect  in  ac- 
cordance with  our  definition  of  equivalent  magnitudes  as 
such  as  can  be  cut  into  pieces  congruent  in  pairs.  Hence  we 
assume-: 

[i]  No  arc  is  less  than  its  chord. 

[2]  No  arc  is  greater  than  the  sum  of  the  tangents  at  its 
extremities. 

As  a  consequence  of  these  paradoxical  assumptions,  an 
approximate  value  of  a  semicircle  is  given  by  the  semiperime- 
ter  of  every  polygon  inscribed  or  circumscribed.  Moreover, 
the  semicircle  cannot  be  less  than  the  inscribed  semiperime- 
ter  nor  greater  than  the  circumscribed. 

514.  Calculating  the  length  of  a  side  in  the  regular  inscribed 
and  circumscribed  polygons  of  6,  12,  24,  48,  96,  etc,  sides, 
radius  i*^"",  and  in  each  case  multiplying  the  length  of  one  side 
by  half  the  number  of  sides,  we  get  the  following  table  of 
semiperimeters : 


« 

i»K 

i"^n 

6 

3.0000000 

3.4641016 

12 

3.1058285 

3 

2153903 

24 

3  1326286 

3 

1596599 

48 

3.1393502 

3 

1460862 

96 

3.1410319 

3 

1427146 

192 

3.1415424 

3 

1418730 

384 

3-1415576 

3 

1416627 

768 

3-1415838 

3 

1416101 

1536 

3.1415904 

3 

1415970 

3072 

3.1415921 

3 

1415937 

6144 

3-1415925 

3 

141 5929 

12288 

3.1415926 

3 

141 5927 

393216 

3.1415926535 

3 

1415926537 

i.e.,  6  X  2i« 

I08  ELEMENTARY  SYNTHETIC  GEOMETRY. 

515.  Since  a  regular  polygon  of  any  number  of  sides,  say 
393216,  is  similar  to  any  other  regular  polygon  of  that  number 
of  sides,  therefore  their  sides  have  the  same  ratio  as  the  radii 
of  their  circum-circles,  or  their  in-circles.  So  3.141592653  is  not 
only  an  expression,  exact  to  nine  places  of  decimals,  for  the 
length  of  the  semicircle  whose  radius  is  i*^",  but  also  for  the 
ratio  of  any  semicircle  to  its  radius. 

516.  This  ratio  of  any  circle  to  its  diameter  Euler  desig- 
nated by  7t. 

The  Bible  [i  Kings  vii.  23]  gives  for  its  value  3.  The  Egyp- 
tians twenty-two  centuries  before  Christ  gave  [4/3]*  =  3- 16. 
Archimedes,  from  the  perimeters  of  the  regular  inscribed  and 
circumscribed  polygons  of  96  sides,  placed  it  between  31^  and 
3|.  Ptolemy  used  n  =  |^^  =  3.1416.  The  Hindoos  gave 
3927/1250  =  3.1416. 

Adriaan  Anthoniszoon,  father  of  Adriaan  Metius  [before 
1589]  gave  355/113  =  3. 14.15929.  Ludolf  van  Ceulen  [1540- 
1610]  gave  Tt  =  3. 141 59265358979323846264338327950288. 

Lambert  in  1761  demonstrated  the  irrationaHty  of  tt.  In 
June  1882  Professor  Lindemann  proved  that  tt  is  a  transcen- 
dental irrational,  that  is,  rt  cannot  be  a  root  of  a  rational 
algebraic  equation  of  any  degree. 

Hence  the  rectification  of  the  circle  is  proved  insoluble  by 
•compassiand  ruler. 

CIRCULAR   MEASURE  OF  AN  ANGLE. 

517.  When  its  vertex  is  at  the  center  of  the  circle, 

any    ^        its  intercepted  arc       arc  any      ^  _  arc 

st'  ^    ~  semicircle  ~  r;r  '  "  (i/7r)st'^  ~    r' 

st'   ^ 
So,  adopting  as  unit  angle ,   that  is,  the  angle  sub- 
tended at  the  center  of  every  circle  by  the  arc  equal  to  its 


MEN  SURA  TION. 


109 


radius,  and  hence  called  a  radian,  then  the  ratio  of  any  angle 
to  the  radian  equals  the  ratio  of  its  arc  to  the  radius. 

If  u  denote  the  number  of  radians  in  an  angle,  and  /  its 
intercepted  arc,  then  u  =  l/r. 

The  quotient  are/radius,  or  u,  is  called  the  circular  measure 
of  an  angle. 

518.  Since  a  triangle  is  half  the  rectangle  of  either  of  its 
sides  and  the  altitude  to  that  side,  therefore  the  area  of  a 
triangle  is  half  the  product  of  the  length  of  a  side  by  the 
length  of  its  altitude. 

519.  Theorem.  In  any  triangle,  the  square  on  a  side  oppo- 
site  any  acute  angle  is  less  than  the  sum 
of  the  squares  on  the  other  two  sides  by 
twice  the  rectangle  contained  by  either  of 
those  sides  and  a  sect  from  the  foot  of 
that  side's  altitude  to  the  vertex  of  the 
acute  angle. 

Proof.  Let  a,  b,  c  denote  the  lengths  of  the  sides,  and  It 
denote  <^'s  altitude,  and  j  the  sect  from  its  foot  to  the  acute 
angle  A. 

a^  _  h^  =  (b  -jy  z=b^  -  2bj-\-f'  =  b''  -  2bj^  c^  -  h\ 
...  a^^b''-2bj-^c\ 

520.  (Heron.)    If  A  denote  the  area    of  any   triangle    and 

s  —  \\a-^b-\-  c),  then  A  =   \/  s\s  —  d\\s  —  b\  \s  —  c\ 

F  -\-  c"  -  a" 


Proof,  j  = 


2b 


.'.  h'  =  c'  -j'  =  c'  - 


(^'  -f  g'  -  aj 
4b' 


.'.  4/i'b'  =  4b'c'  -  lb'  -\-  c'  -  a')\ 


no  ELEMENTARY  SYNTHETIC  GEOMETRY. 


.'.  2hb  =    V4dV'  -  {b'  -\-c'  -  dj, 


.:  4 A  =    V{2bc  +  b'  -[-  c'  -  a')  {2bc  -  b'  -  c'  -{-  a'), 

A  =  i  i/(a-]-b^c)  {b-\-c-a){a-{-b  —  c)  {a  —  b-{-c). 

521.  The  area  of  a  regular  polygon  is  half  the  product  of 
its  perimeter  by  the  radius  of  the  inscribed  circle. 

For  sects  from  the  center  to  the  vertices  divide  the  polygon 
into  congruent  isosceles  triangles  whose  altitude  is  the  radius, 
r,  of  the  inscribed  circle,  and  the  sum  of  whose  bases  is  the 
perimeter,  p,  of  the  polygon.     .*.  N  =  ap/2. 

522.  The  area  of  any  circle  0=  rV. 

For  if  a  regular  polygon  of  393216  sides  be  circumscribed 
about  the  circle  its  area  is  \rp. 

But  \p  is  rit ;  therefore  its  area  is  rV. 


EXERCISES  ON  BOOK   VI.  Ill 


EXERCISES  ON   BOOK  VI. 

1.  In  a  regular  triangle  the  side  {b)  =  \  perimeter  (/)  =  1/3  circum- 
radius  {R)=  2  1/3  in-radius  (r)  =  '- — -  altitude  {K)  =  -y  3  V's'area  (v). 

2.  The  area  of  a  tangent- polygon  (circum-polygon)  is  half  perimeter 
by  in-radius  (ipr). 

3.  How  many  times  greater  does  a  quad'  become  if  we  magnify  it 
until  a  diagonal  is  tripled  ? 

4.  Lengthening  through  A  the  side  ^  of  a  a  by  r  and  c  by  d,  they  be- 
come diagonals  of  a  symtra  which  is  to  the  A  as  (^  +  c)'  to  dc. 

5.  One  vertex  of  a  ||g'm  and  the  mid  points  of  the  other  two  sides 
determine  a  A.     What  is  its  ratio  to  the  ig'm  ? 

6.  The  squares  of  chords  from  the  same  point  are  as  their  ±  projec- 
tions on  the  diameter  from  that  p't. 

7.  Make  a  sq'  equal  to  ^  a  given  sq'. 

8.  Make  a  sq'  =  |  a  given  sq'. 

9.  If  from  an  ^  or  from  supplemental  ^s  we  cut  +  as,  they  are  as 
the  sq's  on  one  of  th^e~=r-sides^ 

10.  Trisect  a  •!•  A  by  |s. 

11.  Their  cross  divides  the  non-|  sides  of  a  trapezoid  externally,  the 
diagonals  internally,  in  the  ratio  of  the  ||  sides. 

12.  (Circle  of  Apollonius.)  If  a  sect  is  cut  in  a  given  ratio,  and  the 
interior  and  exterior  points  of  division  are  taken  as  ends  of  a  diameter, 
this  circle  contains  the  vertices  of  all  triangles  on  the  given  sect  whose 
other  two  sides  have  the  given  ratio. 

•  13.  If  AD  and  BE  are  altitudes  of  a  ABC,  then  a/i>  =  -A,  /  7^. 

AlJ/   BE 


112  ELEMENTARY  SYNTHETIC  GEOMETRY, 


MISCELLANEOUS  EXERCISES  ON  THE  FIRST  SIX  BOOKS. 

1.  Describe  a  0  having  center  in  a  given  st'  and  containing  two  given 
points. 

2.  A  0  may  be  described  which  shall  contain  two  p'ts,  and  have 
r  =  «;  {a>  \AB). 

3.  a  ■\-  b  ■\-  c  >  2a. 

4.  a  +  b  +  c  <  2a  +  2b. 

5.  If  the  sides  of  a  regular  polygon  be  produced  to  meet,  their  inter- 
section  points  are  the  vertices  of  a  similar  polygon. 

6.  Trisect  a  st'  ^. 

7.  From  a  -I-  a,  cut  a  synitra  with  three  sides  =. 
Hint.  Join  extremities  of  the  two  equal  angle-bisectors. 

8.  Two  external  ^  bi's  of  a  •!•  A  are  ||  to  a  side. 

9.  A  median,  a',  is  >,  =,  <  a,  according  as  ^  ^  is  acute,  r't,  or  ob- 
tuse. 

10.  AS  having  a  r't  4-  common,  and  =  hy's,  have  mid's  of  hy's  on  a 
quadrant. 

11.  The  angles  made  by  productions  of  the  sides  of  a  reg'  pentagon 
are  together  a  st'  4-. 

12.  The  angles  made  by  productions  of  the  sides  of  a  reg'  hex'  are 
together  a  perigon. 

13.  Any  two  llg'ms  on  two  sides  of  a  A  are  together  =  to  a  ||g'm  on 
the  third  side,  whose  consecutive  side  is  =  and  ||  to  the  sect  joining  the 
intersection  of  two  sides  produced  of  the  other  llg'ms  to  their  common 
vertex. 

14.  The  squares  on  the  sides  of  a  A  are  together  triple  the  squares 
on  the  sects  joining  the  vertices  to  the  centroid. 

15.  Triple  the  squares  of  the  sides  of  a  a  is  quadruple  the  squares  on 
the  medians. 

16.  The  sum  of  the  sides  of  a  a  is  greater  than  the  sum  of  its 
medians. 

17.  From  the  vertices,  equal  sects  taken  in  order  on  the  sides  of  a  sq' 
give  the  vertices  of  a  sq'. 

18.  With  a  vertex  on  a  vertex,  inscribe  in  a  sq'  a  reg'  A. 


MISCELLANEOUS  EXERCISES  ON  THE  FIRST  SIX  BOOKS.    1 1 3 

19.  Ilg'ms  inscribed  in  a  IJg'm  have  common  sC. 

20.  If  either  diag'  of  a  ||g'm  be  =  to  a  side,  the  other  diag'  >  any 
side. 

21.  Sects  from  a  point  in  a  diag'  of  a  ||g'm  to  vertices  give  As  =  in 
pairs. 

22.  One  median  of  a  trapezoid  bisects  it. 

23.  Sects  from  any  p't  in  a  ||g'm  to  its  vertices  bisect  it. 

24.  Sects  from  the  mid  p't  of  a  non-||  side  of  a  trapezoid  to  opposite 
vertices  bisect  it. 

25.  Medians  of  a  quad'  bisect. 

26.  The  sum  of  sq's  of  diag's  of  a  trap'  =  sq's  of  non-||  sides  +  two 
rect'  of  I  sides. 

27.  Draw  a  chord  bisected  by  a  given  p't  within  a  given  o. 

28.  Any  chords  which  intersect  on  a  diam',  and  make  =  4-^  with  it, 
are  =. 

29.  Describe  a  o  with  given  r,  center  in  given  st',  and  tan'  to  another 
given  St'. 

30.  The  opposite  sides  of  a  circum-quad'  subtend  suppl'  ^s  at  the 
center. 

31.  HD  produced  to  circum-o  is  doubled. 

32.  In  an  inscribed  even  polygon,  non-consecutive  angles  make  half 
the  angle-sum. 

33.  On  a  given  sect  as  chord  describe  a  segment  which  will  contain  a 
given  4-. 

34.  Find  a  curvilinear  figure  equivalent  to  a  regular  even  polygon, 

35.  In  a  regular  even  polygon,  any  vertex  and  the  center  are  co-st' 
with  another  vertex. 

36.  II  chords  are  sides  of  a  symtra. 

37.  The  sum  of  the  squares  of  the  segments  of  two  1  chords  =  d"^. 

38.  The  ±  projections  of  opposite  p'ts  of  a  ©  on  any  st"  are  on  a  con- 
centric 0. 

39.  If  through  a  p't  on  a  common  chord  pass  two  chords,  their  four 
ejctremities  are  concyclic. 

40.  y^s"  =  kt"  +  y^io". 

41.  U\  d::  d:  if,. 

42.  OA'  -H  OB'  +  OC  =  R  +  r. 

43.  Any  rectangle  is  half  the  rectangle  of  the  diagonals  of  squares  on 
its  sides. 

44.  If  two  =  chords  intersect  they  make  equal  segments  [they  are 
diag's  of  a  symtra]. 


114  ELEMENTARY  SYNTHETIC  GEOMETRY. 

45.  A  II  through  the  center  is  \  perimeter  6f  a  circum-symtra, 
ifi.  b  :  c  ::  1.  ir.  A'  on  c  :  1.  fr.  A'  on  b. 

47.  The  sum  of  the  diag's  of  a  quad'  is  less  than  the  sum  of  any  other 
four  sects  from  a  p't  to  the  vertices. 

48.  Through  a  given  p't  draw  a  st'  on  which  is  from  two  given  p'ts 
shall  be  =. 

49.  The  4-  bi's  of  a  |ig'm  make  a  rectangle. 

50.  From  the  r't  4C.  the  median  and  altitude  of  the  r't  A  contain  2(1 
=  dif  of  the  acute  angles. 

51.  An  angle-bi'  and  median  contain  2(!  =  to  dif  of  other  4-^  of  the  a. 

52.  If  of  the  four  As  into  which  the  diag's  divide  a  quad',  two  oppo- 
site are  =,  it  is  a  trap'. 

53.  If  two  circles  cut,  the  intercepts  on  any  two  ||s  through  the  points 
of  section  are  =. 

54.  Chords  all  drawn  from  a  p't  on  a  0  have  their  mid  p'ts  concyclic. 

55.  If  from  one  common  p't  of  two  equal  intersecting  ©s  as  center  a 
O  be  drawn,  two  of  the  points  in  which  it  cuts  them,  and  their  other 
common  p't,  are  cost'. 

56.  If  two  =  OS  cut,  the  part  of  a  st'  through  a  common  p't  inter- 
cepted between  them  is  bi'd  by  the  O  on  their  common  chord  as  diame- 
ter. 

57.  If  two  OS  are  tangent,  two  st's  through  the  p't  of  contact  inter- 
cept arcs  whose  chords  are  ||. 

58.  If  two  OS  touch  externally,  and  ||  d's  be  drawn,  a  st'  joining  their 
extremities  will  contain  the  p't  of  contact. 

59.  In  a  st'  through  the  center  determine  a  p't  from  which  a  tan' 
shall  be  =  d. 

60.  The  4-  made  by  tan's  from  a  p't  to  a  o  is  double  the  4-  of  chord 
of  contact  and  diam'  through  a  p't  of  contact. 

61.  Through  a  given  p't  to  draw  a  st'  which  shall  make  equal  4-^  with 
two  given  st's. 

62.  From  two  given  p'ts  to  draw  two  =  sects  which  shall  meet  on  a 
given  st'. 

63.  From  two  given  p'ts  on  the  same  side  of  a  given  st'  to  draw  two 
st's  which  shall  cross  on  that  st'  and  make  =  4-^  with  it. 

64.  If  a  tan'  be  ||  to  a  chord,  the  p't  of  contact  will  be  the  mid  p't  of 
the  chord's  arc. 

65.  Of  st's  drawn  from  two  given  p'ts  to  meet  on  a  O,  the  sum  of 
those  two  will  be  least  which  make  =  ^s  with  the  tan'  at  the  point  of 
concourse. 


MISCELLANEOUS  EXERCISES  ON  THE  FIRST  SIX  BOOKS.    1 1 5 

66.  If  two  Os  cut,  and  from  either  common  p't  diam's  be  drawn, 
their  extremities  and  the  other  common  p't  are  co-st'. 

d"].  If  a  0  be  described  on  the  r  of  another  ©  as  d,  any  sect  from  the 
common  p't  to  the  greater  is  bisected  by  the  lesser. 

68.  The  st's  joining  to  the  centre  the  intersections  of  a  tan'  with  two 
II  tan's  are  1. 

69.  St's  from  two  p'ts  in  a  O  to  a  p't  in  a  tan'  make  the  greatest  ^ 
■when  drawn  to  the  p't  of  contact. 

70.  If  any  chord  be  bisected  by  another,  and  produced  to  meet  the 
tan's  drawn  at  the  extremities  of  this  other,  the  parts  between  the  tan's 
-ana  the  0  are  =. 

71.  If  one  chord  bisect  another,  and  tan's  at  the  extremities  of  each 
be  produced  to  meet,  the  join  of  their  points  of  intersection  is  ||  to  the 
bisected  chord. 

72.  If  from  the  extremities  of  a  diameter  chords  be  drawn  intersect- 
ing, two  and  two,  on  a  ±  to  that  d',  the  joins  of  the  extremities  of  the 
pairs  are  concurrent. 

73.  If  from  any  p't  in  the  base  of  •}•  a  st's  making  equal  4-'^  with  the 
base  be  drawn  to  the  sides,  the  as  formed  by  joining  the  intersections 
to  the  opposite  vertices  are  =. 

74.  Which  st'  through  a  given  p't  within  a  given  "4-  will  cut  off  the 
least  A  ? 

75.  The  diag's  of  a  trap'  cross  on  a  median. 

76.  A  st'  bisecting  a  side  of  a  A  is  cut  harmonically  by  the  three  sides 
and  a  ||  to  the  bisected  side  through  the  opposite  vertex. 

"J"].  A  st'  from  a  vertex  of  a  A  is  cut  harmonically  by  the  opposite 
side,  a  median,  and  a  ||  to  either  of  the  other  sides  through  the  opposite 
vertex. 

78.  If  from  the  ends  of  a  side  of  a  A  st's  be  drawn  intersecting  in  the 
altitude  to  that  side,  the  straights  joining  the  points  where  they  cross 
the  other  sides  to  the  foot  of  that  altitude  make  equal  angles  with  it. 

79.  If  from  any  2(^  of  a  rectangle  a  sect  be  drawn  to  a  side,  and  a  ± 
to  it  from  the  adjacent  4-  of  the  quad'  so  formed,  their  rect'  =  the  given 
rect'. 

80.  The  two  spherical  tan's  from  a  p't  to  a  0  are  =. 

81.  If  the  g-lines  joining  the  corresponding  vertices  of  two  A'scon- 
<:ur,  the  crosses  of  opposite  sides  are  colli  near  ;  and  inversely. 

82.  If  a  spherical  quad'  is  inscribed,  and  another  circumscribed  touch- 
ing at  the  vertices  of  the  first,  the  crosses  of  the  opposite  sides  of  these 
•quad's  are  collinear. 


Il6  ELEMENTARY  SYNTHETIC  GEOMETRY. 

[The  crosses  of  the  opposite  sides  of  the  inscribed  are  on  the  diago- 
nals of  the  circumscribed.] 

83.  The  crosses  of  the  sides  of  an  inscr'  a  with  the  spherical  tan's  at 
the  opposite  vertices  are  collinear. 

84.  If  from  the  greater  of  two  sides  of  a  A  a  portion  be  cut  ofT  equal 
to  the  lesser,  the  join  of  the  p't  of  section  and  the  opposite  ^  makes  an 
^  =  ^  dif  of  ^s  adjacent  to  third  side  of  a. 

85.  Every  ©  passing  through  a  given  p't  and  centered  in  a  given  st', 
passes  also  through  another  fixed  p't. 

86.  The  rectangles  of  opposite  sides  are  together  double  a  cyclic 
quad'  whose  diag's  are  ±. 

87.  If  through  the  mid  p't  of  any  chord  two  chords  be  drawn,  the 
joins  of  their  extremities  will  cut  off  equal  sects  on  the  first  chord. 

88.  In  a  r't  a  the  dif  between  the  hy'  and  the  sum  of  the  other  sides 
equals  the  ^  of  the  in- O- 

89.  If  from  the  extremity  of  the  radius  of  its  circum-O  bisecting  one 
side  of  a  A  a  J.  be  drawn  to  the  larger  of  the  other  two  sides,  one  of  the 
segments  made  is  half  the  sum,  the  other  half  the  difference,  of  these 
sides. 

90.  The  center  of  O  touching  semicircles  described  outwardly  on  the 
two  sides  of  a  r't  a  is  the  mid  p't  of  the  hypothenuse. 

91.  An  angle-bi'  of  a  a  cuts  the  circum  0  m  the  center  of  a  O  con- 
taining the  other  two  vertices  and  the  in  center. 

92.  If  from  a  vertex  of  a  A  inscribed  in  a  0  st's  be  drawn  ||  to  the 
tangents  at  the  extremities  of  the  opposite  side,  they  cut  oflf  ~  as. 

93.  The  joins  of  the  vertices  and  the  points  of  contact  of  the  in-o  of 
a  A  concur. 

94.  If  from  the  ends  of  a  side  of  a  square  0s  be  described,  one  with 
the  side,  the  other  with  the  diagonal,  as  radius,  the  lune  formed  equals 
the  square. 

95.  If  the  diam'  of  a  semi-0  be  cut  in  pieces  and  on  ;hem  semi-0s 
be  described,  these  together  equal  the  given  semicircle. 

96.  In  a  given  st'  determine  the  p't  at  which  st's  from  two  given  p'ls 
on  the  same  side  of  it  will  contain  the  greatest  ^  . 

97.  If  the  rectangles  of  the  segments  of  two  intersecting  sects  are 
tqual,  their  extremities  are  concyclic. 

98.  If  two  altitudes  are  equal,  is  the  A  isosceles  ? 

99.  If  two  medians  are  equal,  is  the  A  isosceles? 

100.  A  ABC  ~  A  A'B'C  [where  A'  bisects  a;  B',b;  C,  c.\ 


BOOK   VII. 

MODERN  GEOMETRY. 


CHAPTER   I. 

TRANSVERSALS, 

522.  [Menelaus.]  If  the  sides  of  the 
triangle  ABC,  or  the  sides  produced, 
be  cut  by  any  transversal  in  the  points 
<?,  b,  c,  respectively,  then 

\Ab/bC\Ca/aE\\_Bc/cA\  =  i. 

Inversely,  given  this  relation,  the  points  a,  b,  c  will  be  co- 
straight. 

Proof.  Drsiw  BD  \\  to  AC,  and  meeting  the  transversal  inZ>: 

then  Bc/cA  =  BD/Ab,     and     Ca/aB  =  bC/BD ; 

therefore 

[Ca/aBJ_Bc/cA'\  =  \bC/BD\BD/Ab'\  =  bC/Ab  =  \/{^Ab/bC\ 
Inversely,  if  the  straight  ab  meet  AB  in  c\  then  by  Menelaus, 
lAb/bCWCa/aB-WBc'/c'AI  =  i. 
But  by  hypothesis, 

[Ab/bCllCa/aBjlBc/cA]  =  i. 

Therefore  c  and  c'  coincide. 

"7 


'O- 


Il8  ELEMENTARY  SYNTHETIC  GEOMETRY. 

523.  Corollary.   If  a  traversal  intersects  the  sides  AB,  BC, 
CD,  etc.,  of  any  polygon  in  the  points  a,  b,  c,  etc.,  in  order,  then 

[Aa/aB][Bd/dC\\lCc/cD][nd/dE]  .  .  .  etc.  =  i. 

Proof.  Divide  the  polygon  into  triangles  by  straights 
through  one  vertex,  apply  Menelaus  to  each  triangle,  and  com- 
bine the  results. 

524.  [Ceva.]   If  the  sides  of  triangle  ABC 
are  cut  by  A  O,  BO,  CO  in  a,  b,  c,  then 

lAb/bC^iCa/aB\Bc/cA'\  =  i. 

'b  ^         Inversely,  given  this  relation,  the  straights 

iG.  233.  ^^^  j^^^  Q^  ^jjj  i^g  concurrent. 

Proof.  By  the  transversal  ^/5  to  the  t.AaCy^Q  have  [Mene- 
laus] 

{_Ab/bC\CB/Ba\aO/OA'\  =  1 ; 
and  by  the  transversal  Cc  to  the  AaB, 

[Bc/cA][AO/Oa][aC/CB]  =  i. 

Multiply  these  equations  together. 
Inverse  as  in  Menelaus. 

525.  Corollary.  If  transversals  through  O  from  the  vertices 
of  any  odd  polygon  meet  the  sides  AB,  BC,  CD,  etc.,  in  the 
points  a,  b,  c,  etc.,  in  order,  then 

{_Aa/aB\Bb/bC\Cc/cD\Dd/dE\  .  .  .  etc.  =  i. 

526.  Theorem.  If  any  transversal  cuts  the  sides  of  a  triangle 
and  their  three  intersectors  AO,  BO,  CO,  in  the  points 
A' ,  B',  C,  a',  b' ,  c' ,  respectively,  then 

\A'b'  /b'  C\C'  a'  /a'  B'\B'  c'  /c'  A'^  =  t. 

Proof.  Each  side  forms  a  triangle  with  its  intersector  and 
the  transversal.     Take  the  four  remaining  straights  in  succes- 


TRANSVERSALS. 


119 


Pig.  234. 


sion    for   transversals    to   each    triangle,    applying    Menelaus 
symmetrically,  and  combine  the  twelve  equations. 

527.  Theorem.  If  the  vertices 
of  two  triangles  join  concurrently, 
the  pairs  of  corresponding  sides  in- 
tersect co-straightly,  and  inversely. 

Proof.  Take  be,  ca,  ab,  trans- 
versals respectively  to  the  triangles 
OBC,  OCA,  OAB\  apply  Mene- 
laus, and  the  product  of  the  three 
equations  shows  that  P,  Q,  R  lie  on  a  transversal  to  ABC. 

528.  Corollary.  If  the  vertices  of  any  two  polygons  join 
concurrently,  the  pairs  ot  corresponding  sides  intersect  co- 
straightly. 

529.  The  straight  on  which  the  pairs  of  sides  cross  is  called 
the  axis  of  perspective. 

530.  The  ]_ projection  of  a  point  on  a  sect  is  the  foot  of  the 
perpendicular  from  the  point  to  the  straight  of  the  sect. 

531.  The  J_  projection  of  a  sect  on  a  straight  is  the  piece 
between  the  perpendiculars  dropped  upon  the  straight  from  the 
ends  of  the  sect. 

532.  Theorem.  The  _L  projections  on  the  sides  of  a  triangle 
of  any  point  on  its  circumcircle  are  co-straight. 

[This  straight  is  called  the  Simson's  straight  of  the  triangle 
with  respect  to  the  given  point.] 

Proof.  Let  O  be  any  point  on  circumcircle 
of  A  ABC.  Join  its  _L  projections  GF,  GH. 
Join  OA,  OC.  Since  2^  OGC  and  ^  OHC  are 
r't,  therefore  C,  H,  G,  O  are  concyclic.  Simi- 
larly, G,  B,  F,  O  are  concyclic. 

.'.  •4.  OGF  =  4-  OBF,  inscribed  angles  on 
same  arc  of  circle  OGBF. 

But  ^  OBF—  4  OCA,  being  supplemental  to  ^  OBA. 

.-.  40GH-\-4  0GF=  ^.OGH^  4  0CH=st4.. 


Fig.  235. 


120  ELEMENTARY  SYNTHETIC  GEOMETRY. 

533.  Inverse.  If  the  projections  on  the  sides  of  a  triangle 
of  a  point  be  co-straight,  that  point  is  on  the  triangle's  circum- 
circle. 

Proof.  Let  G,  H,  F,  _L  projections  of  O  on  a,  b,  c  be  co- 
straight.  Since  (9,  C,  H,  G  are  concyclic,  .•.  ^  OCH  =  ^  OGF, 
being  supplements  of  ^  OGH. 

But  -^  OGF  =  ^  OBF,  inscribed  angles  on  same  arc  of 
circle  OGBF 

.:  ^  OCA  +  4-  OB  A  =  s't  ^. 
.'.  O,  C,  A,  B  are  concyclic. 


CHAPTER  II. 

HARMONIC    RANGES   AND    PENCILS. 

534.  A  system  of  co-straight  points  is  called  a  range,  of 
which  the  straight  is  the  bearer. 

535.  A  system  of  concurrent  straights  is  called  a  pencil,  of 
which  the  intersection  point  is  the  vertex  or  the  bearer. 

536.  Straights  all  parallel  form  a  pencil  of  parallels  or  a 
parallel-pencil. 

537.  Thus  straights  with  equal  perpendiculars  from  two 
given  points  form  two  pencils,  one  parallel  to  their  join,  and 
the  other  bisecting  it. 

538.  A  range  and  a  pencil  are  called  perspective  when  each 
point  of  the  range  lies  on  a  straight  of  the  pencil. 

539.  Two  ranges  are  called  perspective  when  their  points 
lie  in  pairs  on  the  straights  of  a  pencil.  The  bearer  of  the 
pencil  is  called  XJae.  perspective-center. 

540.  Two  pencils  are  called  perspective  when  their  straights 
cross  in  pairs  in  the  points  of  a  range.  The  bearer  of  the  range 
is  called  the  projection  axis. 

541.  Ranges  and  pencils  are  CdiW^d  projective  if  they  can  be 
put  in  perspective  position. 

542.  If  A,  B  be  two  points,  and  C,  D,  two  co-straight  with 
them,  be  so  taken  that  AC/BC  —  AD/DB,  then  the  points 
A,  C,  B,  D  form  a  harmonic  range;  C  and  D  are  harmonic 
conjugates  with  respect  to  A  and  B\  AC,  AB,  AD  are  said  to 
be  in  harmonic  progression  ;  and  AB  is  said  to  be  a  harmonic 
mean  between  AC  and  AD. 

Thus  we  have  proved  (479)  that  if  C  and  D  are  harmonic 
conjugates  with  respect  to  A  and  B,  then  A  and  B  are  harmonic 
conjugates  with  respect  to  Cand  D. 

121 


122  ELEMENTARY  SYNTHETIC  GEOMETRY. 

543.  If  A,  C,  B,  D  form  a  harmonic  range,  and  O  be  the  mid 
point  of  AB,  then  OA''  =  OB'  =  OC .  OD. 

For  AD/DB  =  AC/BC. 

.'.  (AD  +  DB)/{AD  -  DB)  =  {AC-{-  CB)/{A C -  CB)\ 

:.  2OD/2OB  =  2OB/2OC,     .'.  OB'  =  OC.  OD. 

544.  Theorem.  If  four  concurrent  straights  cut  any  trans- 
s  versal  in  a  harmonic  range,  they  will  cut 

/  \\\    /  every  transversal  in  a  harmonic  range. 

a'./_\_V^'X  T,  Proof.   Through  B  and  B'  draw  BT 


/         Ti'(c  \b/  X       and  j5' 7^  II  to  AA'S,  and   meeting  SC  in 
\    /  7", ,  T/  and  SD  in  Zj, ,  TJ  .     Then  since 

v(,  BT,  =  BT,,  .'.  B'T/  =  B'TJ. 

F"'-  ^36.  A'C'B'D'  is  a  harmonic  range. 

545.  If  A,  C,  B,  D  is  a  harmonic  range,  5^,  SC.  SB,  SD' 
is  a  harmonic  pencil,  and  5(7,  SD  are  harmonic  conjugates  of 
SA,  SB. 

546.  We  have  shown  that  the  arms  of  any  angle  form  with 
its  internal  bisector  and  its  external  bisector  a  harmonic  pencil. 

I  547.  If,  in  a  harmonic  pencil,  one  element  bisect  the  angle 
between  two  conjugates,  then  it  is  perpendicular  to  its  con- 
jugate. 

548.  If  in  a  harmonic  pencil  one  pair  of  conjugates  be  at 
right  angles,  then  these  are  the  internal  and  external  bisectors 
of  the  angle  between  the  other  pair. 

549.  Theorem.  If  two  harmonic  ranges  are  taken,  one  in 
each  of  two  straights,  and  if  three  of  the  four  projection- 
straights  are  concurrent,  then  so  are  the  four. 

Proof.  For  the  three  concurrent  projection-straights  and 
the  straight  from  their  bearer  through  one  of  the  fourth  points 
form  a  harmonic  pencil ;  so  this  latter  straight  contains  also 
the  other  fourth  point. 

550.  Corollary.  If  two  corresponding  points  coincide  in  the 
cross  of  the  two  straights,  then  one  projection-straight  being 
free,  the  other  three  are  always  concurrent. 


CHAPTER    III. 

PRINCIPLE   OF   DUALITY. 

551.  Not  only  the  sect  joining  two  points,  but  the  whole 
straight,  may  be  called  the  join  of  the  two  points. 

552.  The  point  common  to  two  straights  may  be  called  the 
cross  of  the  two  straights. 

553.  In  a  pencil  consisting  of  straights  through  one  fixed 
point,  any  one  of  the  straights  may  be  called  an  element  of  the 
pencil,  or  a  line  on  the  fixed  point  or  bearer. 

In  this  sense,  we  say  not  only  that  points  may  lie  on  a 
straight,  their  bearer,  but  also  that  straights  may  lie  on  a 
point,  their  bearer,  meaning  that  the  straights  pass  through 
this  point. 

554.  In  most  cases  we  can,  when  one  figure  is  given,  con- 
struct another,  such  that  straights  take  the  place  of  points  in 
the  first,  and  points  the  place  of  straights. 

Thus  from  a  definition  or  a  theorem  we  can  obtain  another 
by  interchanging  point  and  straight,  cross  and  join,  range  and 
pencil,  or  by  similar  interchanges. 

555.  A  figure  regarded  as  consisting  of  a  system  of  straights 
crossing  in  points  will  thus  give  a  figure  which  may  be  regarded 
as  a  system  of  points  joined  by  straights ;  and  in  general  with 
any  figure  coexists  another  having  the  same  genesis  from  these 
elements,  point  and  straight,  but  that  these  elements  are  inter- 
changed. 

Any  descriptive  theorem  or  theorem  of  position  concerning 

123 


124  ELEMENTARY  SYNTHETIC  GEOMETRY, 

one,  thus  gives  rise  to  a  corresponding  theorem  concerning  the 
other  figure, 

556.  Figures  or  theorems  related  in  this  manner  are  called 
reciprocal  figures  or  reciprocal  theorems. 

557.  This  correlation  of  point  and  straight  is  termed  a 
principle  of  duality. 

558.  Each  of  two  descriptive  theorems  so  correlated  is  said 
to  be  tJie  dual  of  the  other ;  and  it  will  be  found  that  if  any 
descriptive  property  is  demonstrated,  its  dual  also  holds. 

559.  Since  capitals  mean  points,  and  two  fix  a  straight, 
their  join  ;  so  small  letters  may  denote  straights,  and  two  will 
fix  a  point,  their  cross. 

Thus  AB  denotes  the  straight  which  is  the  join  of  the 
points  A  and  B  ;  while  ab  denotes  the  point  which  is  the  cross 
of  the  straights  a  and  b. 

560.  In  plane  geometry  to  all  points  on  a  straight  the  recip- 
rocal figure  is  all  straights  on  a  point. 

561.  A  sect,  AB,  is  that  piece  of  a  range  containing  the 
initial  point  A  of  the  sect,  its  final  point  B,  and  all  intermediate 
successive  positions  of  the  generating  point. 

562.  The  figure  reciprocal  to  sect  AB  is  ^  ab,  that  piece  of 
a  pencil  containing  the  initial  straight  a  of  the  angle,  its  final 
straight  b,  and  all  intermediate  positions  of  the  generating 
straight. 

RECIPROCAL  THEOREMS. 

563,.  If  two  harmonic  563'.  If  two  harmonic  pen- 
ranges  are  taken,  one  in  each  cils  are  such  that  the  three 
•of  two  straights,  and  if  three  crosses  of  three  pairs  of  corre- 
of  the  four  projection-straights  sponding  straights  are  co- 
are  concurrent,  then  so  are  the  straight,  then  this  straight 
four.  contains     the     cross    of     the 

fourth  pair. 


PRINCIPLE   OF  DUALITY. 


125 


564,.  If  two  harmonic 
ranges  are  taken  one  in  each 
of  two  straights,  and  two 
corresponding  points  coincide 
in  the  cross  of  the  straights, 
then  the  other  three  projec- 
tion-straights are  concurrent. 


564'.  If  two  straights,  one 
in  each  of  two  harmonic  pen^ 
cils,  are  coincident,  then  the 
three  crosses  of  the  other 
three  pairs  of  straights  are 
costraight. 


CHAPTER   IV. 


COMPLETE   QUADRILATERAL  AND   QUADRANGLE. 


565,.  A  system  of  four 
straights,  no  three  concurrent, 
and  their  six  crosses  is  called  a 
complete  quadrilateral,  or  co7n- 
quad. 

566j.  The  four  straights  are 
called  the  "sides  "  of  the  quad- 
rilateral ;  and  the  six  crosses, 
the  vertices. 

567,.  Two  vertices  which 
•do  not  lie  on  the  same  "  side  " 
are  called  opposite  vertices. 

There  are  three  pairs. 

5681.  The  three  straights 
joining  opposite  vertices  are 
called  diagonal  straights,  and 
the  triangle  formed  by  the 
diagonal  straights  is  called  the 
diagonal  triangle  of  the  com- 
plete quadrilateral. 


565'.  A  system  of  four 
points,  no  three  costraight, 
and  their  six  joins  is  called  a 
quadrangle. 

566'.  The  four  points  are 
called  the:  summits  of  the  quad- 
rangle, and  their  six  joins  the 
"  sides." 

567'.  Two  sides  which  do 
not  pass  through  the  same 
summit  are  called  opposite 
sides. 

There  are  three  pairs. 

568'.  The  three  crosses  of 
opposite  sides  are  called  diag- 
onal points,  and  the  ,  triangle 
determined  by  the  diagonal 
points  is  called  the  diagonal 
triangle  of  the  quadrangle. 

126 


COMPLETE   QUADRILATERAL   AND   QUADRANGLE.       1 2/ 


Fig.  237. 


Fig.  238. 


569,.     In     this     complete  569'.    In    this    quadrangle 

quadrilateral  a,  b,  c,  d,  are  the  a,  B,  C,  D,  are  the  summits, 
sides. 

The  vertices  are  i,  2,  3,  4,  The  sides  are  i,  2,  3,  4,  5^ 

5,  6.     I    and    6   are  opposite  6.     i  and  6  are  opposite  sides, 

vertices.       So    are    2    and    5.  So  are  2  and  5.     Al.so  3  and  4. 
Also  3  and  4- 


Fig.  239. 

570,.  In  the  above  com- 
plete quadrilateral,  \{  f  be  the 
joni  of  I  and  6,  ^  of  2  and  5, 
h  of  3  and  4,  then  fgh  is  the 
diagonal  triangle. 

571,.  Theorem.  In  a  com- 
plete quadrilateral  each  pair 
of  opposite  vertices  forms  with 
two  of  the  angular  points  of 
Ihe  diagonal  triangle  a  har- 
monic ranore. 


Fig.  240. 

570'.  In  the  above  quad- 
rangle if  F  be  the  cross  of  i 
and  6,  6^  of  2  and  5,  .^  of  3 
and  4,  then  FGH  is  the  diag- 
onal triangle. 

571'.  In  a  quadrangle,  each 
pair  of  opposite  sides  forms 
with  two  of  the  sides  of  the 
diagonal  triangle  a  harmonic 
pencil. 


128 


ELEMENTARY  SYNTHETIC  GEOMETRt 

W 


Fig.  241.  Fig.  242. 

Proof.  The  range   q\ASC^^  Proof.  The  pencil  Q\asc]  is. 

is  perspective  with  the  range  perspective    with    the    pencil 

r\^ESF']  to  projection  center  B  ;  R{esf\  to  projection  axis  b ;  on 


.*.  on  a  straight  through  i5must 
lie  the  harmonic  conjugates  R 
and  ^  to  5"  of  the.se  ranges. 

But  also  q\ASC^  is  per- 
spective with  rlFSEA^  to  pro- 
jection center/?;  .-.also  on  a 
straight  through  D  must  lie 
R  and  Q.  Hence  they  must 
lie  on  s,  the  join  of  B  and  D. 


b  must  cross  the  harmonic 
conjugates  r  and  ^  to  .y  of 
these  pencils. 

But  also  0,\asc\  is  projec- 
tive to  R\^fse\  to  projection 
axis  d\  .'.  on  d  also  must  cros? 
r  and  g.  Hence  they  must  gc 
through  S,  the  cross  of  b  and 
d. 


572.  Problem.  To  draw  a  pair  of  tangents  to  a  given  circle 
from  a  given  external  point  by  means  of  a  ruler  only. 

Construction.  From  the  given 
point  S,  draw  SCA,  SEF,  cutting  the 
given  circle  in  A,  C  and  E,  F.  Join 
AE,  CF,  crossing  at  B.  Join  AF, 
CE,  and  produce  to  meet  at  D. 
The  st'  DB  contains  the  chord  of 
contact  of  S. 

For  we  have  proved  in  [497]  that 
the  chord  of  contact  of  5  contains 
the  harmonic  conjugates  R,  Q  of 
S  on  EF 2iWdiAC,  and  we  have  just 
proved  in  [571,]  the  opposite  ver- 
tices BD  of  the  complete  quadrilateral  abed  costraight  with 
R,  Q,  these  harmonic  conjugates  of  5. 


,<v 


v^/ 


CHAPTER  V. 


INVERSION. 


573.  If  on  a  ray  from  a  fixed  point  O  we  take  /*,  and  F' 
such  that  the  rectange  OP, .  OP'  equals  the  square  on  a  fixed 
sect  r,  then  the  points  /*,  and  P'  are  termed  each  the  inverse 
of  the  other  with  regard  to  O,  the  center  of  inversion,  and  r,  the 
radius  of  invtirsion. 

574.  Any  two  points  and  their  inverses  are  concyclic. 

575.  If  /*,  moves  on  a  certain  hne,  then  P'  describes  that 
line's  inverse. 

576.  Theorem.  The  inverse  of  a  circle  through  (9  is  a 
straight  perpendicular  to  the  diameter  through  O.  <^r  i 

Proof.  Let  A,  be  the  other  end  of  the  diameter  through 
O,  and  /*,  any  other  point  on  the  circle.  Take  /"  and  A' 
such  that  OP,  .OP'  =  OA,.  OA'  =  r\  Then  A,,  A',  P,,  P' 
are  concycHc,  .*.  ^  OA'F  —  ^  OP,A^ ,  .'.  P'  is  on  the  perpen- 
dicular to  OA'  through  the  fixed  point  A'. 

577.  Corollary.  If  the  straight  is  tangent  to  the  circle, 
the  center  of  inversion  is  the  other  end  of  the  diameter 
through  the  point  of  contact,  and  this  diameter  is  the  radius 
of  inversion. 

If  the  straight  cuts  the  circle,  either  end  of  the  diameter  J, 

129 


I30 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


to  it  may  be  taken  as  the  center  of  inversion,  the  radius  of  in- 
version being  the  sect  from  this  to  either  point  of  section. 
Thus  the  circum-circle  of  an  isosceles  triangle  can  be  inverted 
into  the  straight  through  the  equal  angles. 

578.  Theorem.  The  inverse  of  a  circle  not  through  O  is 
another  circle. 

Proof.  Draw  OAB  through  the  center  of  the  given  circle. 
Take  the  inverse  points  of  A^,  B^,  P^.  Then  OP^.  OP'  =z 
OA,.OA'.  .-.  A,,  A',  P,,  P'  are  concyclic,  .-.  ^  OA'F  = 
^  OP,A,.  In  the  same  way  ^  A'B'F  =  ^  B,P,P',  .-.  ^  OA'F 
-f  yf  A'B'P'  =  ^  OP,A,  +  -4.  B,P,P'  =  r't  ^  (since  ^  A,P,B, 
is  r't),  .-.  4  A'P'B'  is  r't     .-.  P'  is  on  ©  with  diamr*  A'B'. 


Fig.  24s. 

LINKAGE. 
579.  The  Peaucellier  Cell  consists  of  a  rhombus  movably 


Fig.  246, 


jointed,  and  two  equal  links  movably  pivoted  at  a  fixed  point, 
and  at  two  opposite  extremities  of  the  rhombus. 


INVERSION. 


131 


TO   DRAW    A   STRAIGHT   LINE. 


580.  Take  an  extra  link,  and,  while  one  extremity  is  on  the 
fixed  point  of  the  cell,  pivot  the  other  extreniity  to  a  fixed 
point.     Then  pivot  the  first  end  to  one  of  the  free  angles  of 


Fig.  247. 


the  rhombus.  The  opposite  vertex  of  the  rhombus  will  now 
describe  a  straight  line,  however  the  linkage  be  pushed  or 
moved. 

Proof.  By  the  bar  FD  the  point  D  is  constrained  to  move 
on  the  circle  ADR.  A,  D,  E  are  always  on  the  r't  bi'  of  BC. 
Therefore,  if  AE .  AD  is  constant,  E  moves  on  the  straight 
EM.  ButAE.AD=[AN-\-NE']\_AN-NE]  =  AN'-NE' 
=  [AN'  +  NR^  -  \_NE^  +  NE"^  =  AB'  -  BE""  =  2.  constant. 


Fig.  248. 


132  ELEMENTARY  SYNTHETIC  GEOMETRY. 

581.  A  linkage  called  Hart's  Contraparallelogram  is  formed 
of  four  links  ^^  =  67?  sides,  and  ^C  =  ^Z>  diagonals  of  a  sym- 

c_    _ivi B  tra.    The  mid  points  O,  P^ ,  P,  O' 

.   /"^X' /  '^v.,^-^^  ^''^    always    costraight,    and    the 

Oy^'"'   Pj>^^  p'       \  rectangle      OP^  .  OP',     constant. 

//    >\  \    /  ^^"---..N.        For  if   M,  H  be   the    mid  points 

D^^ -"h ^-^A    of     the       symtra's    ||    sides,    then 

Fig.  249-  OM  =  OH  ^nd    MP,  =  MP'  = 

HP'  =  HP, .  [The  sect  joining  the  mid  points  of  two  sides  of 
a  triangle  is  half  the  third]  ;  so  the  relative  position  of  the 
points  OP,P'  is  kept  the  same  as  in  Peaucellier's  Cell.  So  if  O 
is  fixed  and  P,  describes  any  line,  then  P'  must  describe  its 
inverse. 


CHAPTER  VI. 

POLE  AND   POLAR  WITH   RESPECT  TO  A  CIRCLE. 

582.  Theorem.  If  the  cross  of  two  tangents  glides  on 
a  straight,  their  chord  of  contact  rotates  about  a  point ;  and 
inversely. 


Fig.  250. 


Proof.  Draw  CR  _L  to  /,  meeting  TT'  \x\  P.  Since  CG  is 
r't  bi'  of  TT,  .'.  ^  PDG  is  r't.  .-.  P,  R,  D,  G  are  concyclic  ; 
.-.  CR .CP  =  CD.CG  =  CT'  [since  TD  is  ±  to  hy'  of  r't 
A  CTG].  But  CR  and  CT dixe.  fixed;  .'.  also  CP.  Inversely, 
draw  GR  _L  to  CP.  Since,  in  the  inverse,  CP  and  CT  are 
fixed,  .'.  so  also  is  CR. 

P  is  called  the  pole  of  p,  and  /  the  polar  of  P  with  respect 
to  the  given  circle. 

583.  Since  R  and  P  are  inverse  points,  with  respect  to  the 
center  C,  and  radius  CT,  therefore  the  perpendicular  to  their 

133 


134  ELEMENTARY  SYNTHETIC  GEOMETRY. 

Straight  through  either  of  two  inverse  points  is  the  polar  of  the 
other,  which  is  the  pole  of  the  perpendicular. 

584.  To  get  the  pole  of  P  with  respect  to  a  circle  with 
center  C,  join  PC  cutting  the  circle  in  A  and  B,  and  through 
R,  the  harmonic  conjugate  of  P  with  regard  to  AB,  draw  a 
perpendicular/. 

585.  Inversely,  the  pole  of/  with  respect  to  a  circle,  center 
C,  is  the  harmonic  conjugate  of  the  foot  of  the  perpendicular 
from  C  on  p  with  regard  to  the  intersection  points  of  the  per- 
pendicular with  the  circle. 

586,.  If  a   straight   passes  586'.   If  a  point  lie  on  the 

through  the  pole  of  a  second  polar   of   a   second    point,   so 

straight,   so    does  the  second  does  the  second  point  lie  on 

straight  pass  through  the  pole  the  polar  of  the  first, 
of  the  first. 

A    a' 

Proof.  If  A  lie  on  AR  the  polar  of  B, 
then  CB  _L  to  AR.  Draw  BD  J_  to  CA, 
then  A,  R,  B,  D  are  concyclic ; 

.-.  CA  .  CD  =  CR.CB  =  r'',  .'.  BD 
is  the  polar  of  A. 

Fig.  252. 

587,.  Corollary.    The   join  587'.  The  cross  of  the  polars 

of  the  poles  of  two  straights  is      of  two  given  points  is  the  pole 
the  polar  of  their  cross.  of  their  join. 

588.  Theorem.  Any  secant  through  a  pole  is  cut  harmoni- 
cally by  the  circle,  pole,  and  polar. 

We  have  proved  this  in  [497]  if  the  pole  be  without  the 
circle.  If  the  pole  be  within  the  circle,  we  may  substitute  for 
it  the  intersection  of  its  polar  and  the  secant,  since  [586]  the 
polar  of  that  point  contains  the  given  pole. 

589.  A  triangle  of  which  each  side  is  the  polar  of  the  oppo- 
site vertex  with  regard  to  a  circle  is  said  to  be  self -conjugate 
with  respect  to  the  circle. 


POLE  AND  POLAR    WITH  RESPECT   TO  A    CIRCLE.       1 35 


590.  The  diagonal  triangle  of  a  quadrangle  inscribed  in  a 
circle  is  self-conjugate. 

Proof.  Ranges  QAD,  QBC  are 
perspective  from  center  S\  .'.  the 
harmonic  conjugates  to  Q  are  on 
a  straight  through  5.  But  ranges 
QAD,  QCB  are  perspective  from 
center  R  ;  .".  the  harmonic  con- 
jugates to  Q  are  on  a  straight 
through  R.  .'.  SR  is  the  polar 
of  Q.  In  the  same  way  QR  is  the  polar  of  S. 
polar  of  R. 

591.  Corollary.  With  ruler  only,  draw  the  polar  of  a  given 
point,  or  find  the  pole  of  a  given  straight,  with  respect  to  a 
given  circle. 


Fig.  253. 


OS  is  the 


Fig.  2SS. 

592,.    The   polars    of   the  592'.  The  poles  of  the  four 

four  points  of  a  harmonic  straights  of  a  harmonic  pencil 
range  form  a  harmonic  pencil,      form  a  harmonic  range.  ' 

Proof.  Let  P  be  the  pole  of  the  straight  ACBD  with 
respect  to  ©a  Oi  A,  C,  B,  D,  the  polars  PA',  PC,  PB' , PD' , 
are  _L  to  OA,  OC,  OB,  OD.  Thus  the  angles  between  the 
straights  PA' ,  PC,  PB',  PD'  are  respectively  equal  to  the 
angles  of  the  harmonic  pencil  OA,  OC,  OB,  OD. 

593.  If  with  respect  to  a  given   fixed   circle   be  taken  the 


136 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


pole  of  each  straight,  the  polar  of  each  point,  of  a  figure  F^,  we 
obtain  a  dual  figure  F' .  This  method  is  called  polarization  or 
reciprocation,  and  either  of  the  figures  is  termed  the  polar 
reciprocal  of  the  other,  and  any  geometrical  property  of  the 
one  has  its  correlative  for  the  other. 

594j.    The    pole    of   each  594'.    The 

straight   through  a  point  lies     point     on     a 


each 


on  the  polar  of  this  point. 

595,.  The  join  of  the  poles 
of  two  straights  is  the  polar  of 
their  cross. 

596,.  The  poles  of  the 
straights  of  a  pencil  form  a 
range  whose  bearer  is  the 
polar  of  the  pencil's  vertex. 


the 


polar    of 
straight     goes 
pole     of     this 


cross    of 
points    is 


the 
the 


through 
straight. 

595'.  The 
polars  of  two 
pole  of  their  join. 

596'.  The  polars  of  the 
points  of  a  range  form  a  pen- 
cil whose  vertex  is  the  pole  of 
the  range's  bearer. 


597.  Thus  in  reciprocal  polars  correspond 
in  /^ 


a  jom, 
a  pencil, 
parallels, 

angle  between  two  straights; 


\xvF' 
a  cross, 
a  range, 

points  co-straight  with  the 
center  of  reciprocation, 

^  [or  its  supplement]  sub- 
tended by  two  points  at  center 
of  reciprocation ; 
and  vice  versa. 

598.  A  self-conjugate  triangle  is  its  own  reciprocal  polar. 

599.  The  diagonal  triangle  of  a  cyclic  quadrangle  is  also 
the  diagonal  triangle  of  the  complete  quadrilateral  whose  sides 
touch  the  circle  at  the  vertices  of  the  quadrangle. 


Fig.  256. 

CHAPTER  VII. 

CROSS   RATIO. 

600.  If  in  a  range  consisting  of  four  points,  A,  B,  C,  D,  we 
take  A  and  B,  called  conjugate  points,  as  the  extremities  of 
a  sect,  this  is  divided  internally  or  externally  by  C ;  and  dis- 
tinguishing the  "  step''  AC  from  CA  as  of  opposite  "  sense,"  so 
Vhat  AC  ^  —  CA,  the  ratio  AC/BC  is  never  the  same  for  two 
positions  of  C.  The  like  is  true  of  the  positive  or  negative 
number  AD/BD. 

137 


N 


R. 


:u2 


ii^ 


m  '  73]) 


138 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


_JBJ5 

D..    1,     -Tl 


The  ratio  [^(7^(r]/[^Z>/^Z>]  is  called  the  cr^j^  r«//^  of 

.  . » ■  ■  — ^ — """ 

the  range,  and  is  written  {^ABf.CD\ 

601.    Four    elements    may     be    arranged    in    twenty-four 

different  ways: 

^      {_AbCD\  \bAdC\  \CbAB\  {_DCBA\ 

\     [ABDCl  [BACD],  [DCABl  [CDBAl 

^    [[ACBD],  [CADB],  [BDAC],  [DBCA], 

\,\[aCDBI  \CABD\  IDBAC\  \_BDCA\ 

y    lADBC],  ibACBl  IBCAD\  {CBDA\ 

'    \^ADCB\  \_bABC\  \CBAD\  \BCDA\, 


ft;5 


but  four  cross  ratios  in  each  of  these  six  rows  are  equal,  as  may 
be  readily  proved  by  writing  out  any  two  in  a  row. 

602.  If  in  a  cross  ratio  the  two  points  belonging  to  one  of 
the  two  groups  be  interchanged,  the  cross  ratio  changes  to  its 
reciprocal. 

[Proved  by  writing  out  their  values.] 

Thus  the  ratio  in  the  second  row  is  reciprocal  to  that  in  the 
tirst,  fourth  to  third,  sixth  to  fifth. 

603.  If  in  a  cross  ratio  the  two  middle  letters  be  inter- 
changed, the  cross  ratio  changes  to  its  complement. 

{^ABCD'\  =  I  -  [ACBDl 

For  we  have,  taking  account  of  sense  or  sign,  BC  -\-  C  A-^- 
AB^o; 

BC.  AD  -^  CA  .  AD  -^  AB .  AD  =  o; 
BC.AD^CA.[_BD+AB^+AB.ICD-CA]  =^0; 
BC.AD-{-CA.BD-\-  AB.CD  =  o; 
I  +  [CA  .BDyiBC.AD]  +  [AB.  CDyiBCAD]  =  0; 
I  -[AB.  CD]/[CB.  AD]  =  IAC.  BD]/[BC.  AD]  ; 


;/■ 


AB  /AD  __AC/Ap 
~CD  "  BC'  Bb ' 


CB 
I  -  \ACBD] 


__AC  1. 
"Bc'  ' 

[AbCD].. 


CHOSS  RATIO. 


139 


604.  By  603    [AbCEX  =   I  -  [ACDB]  =  [by   602]    i  - 
l/[ACBD]  =  [by  603]  I  -  i/[i_-  {ABCno 

Thus  if  the  cross  ratio  {ABCB]  =  A,  then   the  six  cross 

I 


ratios  derivable   from   these    four  co-straight  points  are  A, 
A  -  I        A 


A' 


^  ~  '^'  I  -  A'     A  '  A  -  r 

605.  Theorem.  If  5  be  a  point  without  the  range  ABCD, 
and  if  through  C  a  straight  be  drawn  parallel  to  SD,  meeting 
SA,  SB  in  G,  H,  respectively,  then  GC/HC-  [ABCD\ 


Fig.  257. 


Proof.  GC/SD  =  CA/DA.     SD/HC=DB/CB. 

GC  Sp  _CA    DB      CA    iCB 
SD'  HC 

.:  GC/HC 


I  LB 
'DA'~CB  ~DA/^B' 

ICDAB^  =  [ABCD]. 


606.  If    two    transversals     meet     the 
straights  of  the  pencil  5  \^abcd\  in  A,  B,C,  a 
D  and    in  A',  B\  C,  D\  then  [AbCD]        ;g 
=  [A'B'CB']. 

Proof.  Through  C  and  C  draw  GH  and  ^ 
G'H'  II  to  SD.  Then ,  GC/HC  =  G'  C'/H  C.  •    njh 

Fig.  258. 

607.  The  cross  ratio  of  the  pencil  5  [abed']  means  the 
cross  ratio  of  the  four  points  ABCD  on  any  transversal,  and  is. 
written  5  [ABCD]. 


140 


ELEMENTARY   SYNTHETIC  GEOMETRY. 


608.  If  two  ranges  or  pencils  have  equal  cross  ratios  they 
■are  said  to  be  eqiii-cross. 

609.  Mutually  equiangular  pencils  are  equi-cross. 


Fig.  260. 


610'.  The  crosses  of  cor- 
responding straights  of  two 
equi-cross  pencils  which  have 
two  corresponding  straights 
coincident  are  co-straight. 


Fig.  259. 

610,.  The  joins  of  corre- 
sponding points  of  two  equi- 
cross  ranges  which  have  two 
corresponding  points  coinci- 
dent are  concurrent. 

Proof.  Let  the  join  of  the  two  crosses  B  and  C  cut  the 
common  straight  in  A,  and  cut  d  in  D.  Then  is  D  also  on  d' , 
since  by  hypothesis  d'  cuts  ABC  in  a  point  D'  such  that 
IAbCD'\  =  [ABCDf]. 

611.  Corollary.  Equi-cross  ranges  or  pencils  are  projective. 

612,.  Pencils  whose  straights  612'.   Ranges  whose  points 

pass  through  four  fixed  points     lie  on  four  fixed  tangents  to  a 


on  a  circle,  and  whose  vertices 
lie  on  the  circle,  are  equi-cross. 

Proof.  The  pencils  are  mu- 
tually equiangular. 

6i3i.  [Pascal.]  In  a  cyclic 
hexagon  the  crosses  of  oppo- 
site sides  are  co-straight. 


circle  and  whose  bearers  are 
tangent  to  the  circle  are  equi- 
cross. 

Proof.  Polarization  from 
612,. 

613'.  [Brianchon.]  In  a  cir- 
cumscribed hexagon  the  joins 
of  opposite  vertices  are  con- 
current. 


CROSS  RATIO. 


I4I: 


Fig.  261. 


Proof.  By6i2„  the  pencils  B.ACDE,  and  F.ACDE  are 
equi-cross ;  .-.  iRHDE]  =  VQCDL]  ;  .*.  by  (610,)  RQ,  HQ  EL 
are  concurrent. 

614.  If  the  figure  formed  by  joining  the  six  concyclic  points, 
by  consecutive  sects  in  any  order  be  called  a  hexagram,  there 
are  60,  and  Pascal  holds  for  each. 

615.  Pascal  holds  for  six  points,  three  co-straight  and  also, 
the  other  three. 


INVOLUTION. 

616.  If  a  system  of  pairs  of  co-straight  points  AA\  BB\ 
CC,  etc.,  be  so  situated  with  regard  to  a  point,  O  on  the  same, 
straight  that  OA  .  OA'  =  OB .  OB'  =  OC .  OC,  etc.,  they  are 
said  to  be  in  invohition.  The  point  O  is  called  the  center,  and 
AA' ,  BB',  CC,  etc.,  are  called  conjugate  points  of  the  involution. 

The  points  E,  F,  situated  on  the  range,  on  opposite  sides  of 
C,  such  that  OE"  =  OF'  =  OA  .  OA'  are  called  the  double  points 
of  the  involution. 

If  straights  be  drawn  from  a  point  .S  outside  the  range  to 
A,  A',  B,  B',  C,  C,  etc.,  they  form  a  pencil  in  involution,  and 
SE,  SF  are  called  the  double  straights  of  the  pencil. 

[Observing  sense  or  sign,  the  double  points  and  double 
straights  are  real  only  when  conjugate  points  of  the  involution; 
are  on  the  same  side  of  the  center.] 


142  ELEMENTARY  SYNTHETIC  GEOMETRY. 

617.  Theorem.  The  two  double  points  and  any  pair  of  con- 
jugate points  of  an  involution  form  a  harmonic  range. 

For  the  sect  between  two  double  points  is  diameter  of  the 
circle  with  regard  to  which  the  conjugate  points  are  inverses. 

618.  In  a  system  of  points  or  straights  in  involution  the 
cross  ratio  of  any  four  points  or  straights  is  equal  to  that  of 
their  conjugates. 

619.  If  two  pairs  of  conjugate  straights  of  a  pencil  in  invo- 
lution be  at  right  angles,  then  every  pair  of  conjugate  straights 
are  at  right  angles. 

RADICAL  AXIS. 

620.  Corollary.  The  straights  of  a  series  of  right  angles  at 
the  same  vertex  form  a  system  in  involution. 

621.  Points  from  which  tangents  to  two  given  circles  are 
equal  lie  on  a  perpendicular  to  the  center-sect  which  so  divides 
it  that  the  difference  of  the  squares  on  the  segments  is  equal 
to  the  difference  of  the  squares  on  the  radii  of  the  two  circles. 

622.  The  bearer  of  the  points  from  each  of  which  tangents 
drawn  to  two  given  circles  are  equal  is  called  the  radical  axis 
of  the  two  circles. 

623.  If  two  circles  intersect,  their  radical  axis  contains  their 
common  chord. 

MILNE'S   SYMMETRY  THEORY   OF   MAXIMUM   AND   MINIMUM. 

623,.  If  a  continuously  varying  magnitude,  changing  in 
accordance  with  some  definite  law,  first  increases  until  it  attains 
a  certain  value  and  then  decreases,  that  specific  value,  both 
preceded  and  followed  by  lesser  values,  is  called  a  inaxiimiin 
value  of  the  varying  quantity. 

Similarly,  a  value  immediately  preceded  and  followed  by 
greater  values  of  the  variable  is  called  a  minimum  value. 

623,.  Just  so  the  form  of  a  geometrical  figure,  varying  in  a 
definite  way,  may  approach  symmetry,  may  attain  symmetry, 
may  immediately  become  unsymmetrical. 

6233 .  The  positions  which  give  the  maximum  and  minimum 


CROSS  RATIO.  143 

values  of  a  continuously  varying  geometrical  magnitude  in  any 
figure  are  positions  of  symmetry  with  regard  to  other  parts  of 
the  figure  which  are  fixed  in  position. 

For  example,  the  _L  from  a  chord  to  its  arc  is  a  maximum 
when  on  the  axis  of  symmetry  of  the  figure. 

Again,  the  perimeter  of  a  triangle  of  fixed  surface  on  a 
given  base  is  a  minimum  when  the  A  is  -l- . 

623^.  Thus  every  varying  geometrical  magnitude  may  be 
considered  to  have  two  properties,  one  metrical,  one  positional 
or  descriptive. 

When  the  magnitude  has  a  symmetrical  position  it  has  a 
maximum  [minimum]  value,  and  inversely.  So  we  may  reduce 
the  problem  of  finding  the  maximum  [min']  values  of  any 
varying  geometrical  quantity  to  the  much  simpler  one  of  find- 
ing its  positions  of  symmetry. 

Ex.  I.  The  minimum  [max']  sect  between  two  Os  is  on 
their  axis  of  -I-  [is  on  their  center-st']. 

Ex.  2.  A  and  B  are  two  fixed  p'ts  without  a  given  O  O. 
Find  a  p't  P  on  the  o  such  that  AF"  -\-  BP^  may  be  a  mini- 
mum. 

Bisect  AB  in  C.     Then  AP^  -[-  BP''  =  2AO^  iCP^ . 

Now  AC  is  constant,  .*.  CP  must  be  a  minimum  ;  .-.  the  re- 
quired p't  is  where  CO  cuts  the  ©.  [For  max'  take  its  other 
-cross.] 

Ex.  3.  In  Ex'  2  substitute  a  st'  for  the  Q. 

Ex.  4.  Through  a  given  p't  draw  a  chord  which  shall  cut 
off  a  minimum  surface. 

[The  p't  must  be  on  the  -I-  axis,  .-.  it  bisects  the  chord.] 

Ex.  5.  Substitute  in  Ex'  4  two  intersecting  st's  for  the  arc. 

[Same  solution.] 

Ex.  6.  Through  a  given  p't  within  a  Q  draw  the  minimum 
■chord.    [Same  solution.] 

Ex,  7.  If  two  sides  of  a  a  be  given  in  magnitude,  the  sur- 


c     ■'   >' 


144  ELEMENTARY  SYNTHETIC  GEOMETRY. 

face  is  a  maximum  when  each  is  -I-  axis  for  the  st'  of  the  other 
[when  they  are  _L]. 

Ex.  8.  To  cut  a  sect  so  that  the  rectangle  of  the  2  pieces^ 
may  be  a  maximum. 

[The  mid  p't.] 

[For  the  same  p't,  the  sum  of  the  sq's  on  the  segments  is 
min.] 

Ex.  9.  The  p't  within  a  sq'  such  that  sq's  on  the  _Ls  from  it 
to  the  sides  are  together  a  minimum  is  its  symcenter. 

Ex,  10.  If  two  sects  cut  J_,  the  sum  of  the  rectangles  of  the 
segments  of  each  is  a  maximum  when  they  mutually  bisect. 

Ex.  II.  In  a  given  square  inscribe  the  minimum  sq. 

Ex.  12.  The  p't  within  a  a  such  that  the  sum  of  the 
squares  of  its  sects  from  the  vertices  is  a  minimum,  is  the 
centroid. 

Ex.  13.  Within  a  A  find  a  p't  such  that  the  sum  of  the 
squares  on  the  _Ls  from  it  to  the  sides  may  be  a  minimum. 


fA  _ A _ A  _  2A \ 

\a~  b~  c~  a^  A-b"  A-  c'J' 


[For  development  of  this  theory  see  Milne's  Companion  to 
Problem  Papers.] 


EXERCISES  ON  BOOK   VII.  I45 


EXERCISES  ON   BOOK   VII. 

1.  The  Simson-line  of  a  p't  bisects  the  join  of  that  p't  and  the  or- 
thocenter  of  the  A. 

2.  Circles  described  on  any  3  chords  from  one  p't  of  a  O  as  diameters 
have  their  other  3  p'ts  of  intersection  co-straight. 

3.  The  circum-Os  of  the  4  As  formed  by  4  intersecting  straights 
concur. 

4.  The  diameter  of  the  in-o  of  a  r't  A  and  the  hypothenuse  together 
equal  the  other  sides. 

5.  If  A,  B,  C,  D  are  concyclic,  show  that  the  Simson  linesof /?,  B,  C,  D 
with  respect  to  as  BCD,  CDA,  DAB,  ABC,  and  the  nine-point-circles 
of  those  AS,  all  pass  through  the  same  point. 

6.  The  bisectors  of  the  ^  s  in  a  segment  of  a  O  form  2  pencils,  whose 
bearers  are  the  ends  of  the  diameter  bisecting  the  segment. 

7.  If  from  a  p't  within  a  A  is  be  drawn  to  the  sides,  then  is  the  sum 
of  the  sq's  of  the  three  segments  of  the  sides  which  have  no  common 
end  point  equal  to  the  sum  of  the  squares  of  the  other  3. 

[True  when  the  p't  is  on  or  without  the  perimeter  of  the  A.] 

8.  Inverse  of  the  preceding. 

9.  If  from  a  p't  within  a  A  ±s  be  drawn  to  the  sides,  then  is  the  sum 
of  the  3  rectangles  of  side-segments  having  no  common  end  p't  each 
with  its  A-side  equal  the  sum  for  the  other  3,  and  equal  half  the  sum  of 
the  squares  of  the  sides  of  the  A. 

10.  The  3  internal  and  3  external  bisectors  of  the  ^s  of  a  a  meet  the 
opposite  sides  in  6  p'ts,  which  are  3  by  3  in  4  st's. 

11.  If  of  4  p'ts  one  is  the  orthocenter  of  the  other  3,  then  every  one  is 
the  orthocenter  of  the  other  3. 

12.  A,  B,  Cand  their  orthocenter //"are  the  centers  of  the  4  ©s  which 
touch  DEF,  the  orthocentric  a. 


-\  1 


BOOK    VIII. 

RECENT  GEOMETRY, 
[The  Lemoine-Brocard  Geometry.] 
CHAPTER   I. 

ANTI-PARALLELS,  ISOGONALS,  SYMMEDIANS. 

624.  Two  mutually  equiangular  polygons  are  co-sensal  when 
rays  pivoted  within  them  and  containing  the  vertices  of  equal 
angles,  rotate  in  the  same  sense  to  pass  through  the  vertices  of 
the  consecutive  equal  angles. 

625.  If  two  transversals  cross  an  ^ ,  so  as  to  make  with  its 
arms  two  As  equiangular,  but  not  co-sensal,  then  either  of 
these  transversals  is  said  to  be  anti-parallel  to  the  other  with 
regard  to  the  angle. 

Thus,  if  ABK,  ACHhe  straights,  and  ^  AKH  -  4  ACB, 
then  KH  anti-|  to  BC  with  respect  to  :2f  A. 

626.  KB  and  HC  are  anti-||  with  regard  to  the 
■4-  of  BC  with  KH. 

627.  B,  C  H,  K  are  concyclic ;  and  inversely, 
if  a  quad'  is  cyclic,  either  opposite  pair  of  its  sides 
are  anti-||  with  regard  to  the  ^  between  the  othei 
pair.  • 

628.  Any  St'  II  to  the  tan'  to  circum-Q  of   A 
ABC  at  A  is  anti-||  to  BC 

629.  Anti-||s  to  2  sides  of  a  a  make  the  same  ^s  with 
the  3d  side. 

Thus  anti-IJs  to  the  sides  a  and  /5  of    A  ABC  make  each 

with  <:  an  ^  =  C. 

146 


Fig.  263. 


ANTI-PARALLELS,    ISOGOXALS,    SYMMEDIANS. 


^A7 


If  the  join  of  their  2  ends  not  in  the 
3d  side  is  |  to  it,  they  are  =,  since  their  4 
ends  are  then  vertices  of  a  symtra. 

Inversely,  if   2  anti-Ils  are  =,  their  4 

B'  A  . 

Fig.  264.  ends  are  vertices  of  a  symtra. 

In  each  case  the  center  of  the  circle  circumscribing  the 
symtra  is  in  the  bisector  of  the  ^  between  the  anti-fs. 

630.  The  joins  of  the  feet  D,  E,  F  of  the  altitudes  of  a  a 
ABC  are  anti-jl  to  its  sides.  A  ABC  is  called  the  original  tri- 
angle, and  A  DEF  is  called  the  orthocentric  triangle. 

631.  Given  any  anti-!|  E'F \.o  a  within  A  ABC;  two  anti-[|', 
FD,  D'E,  to  b,  c,  can  always  be  found  within  the  A.  [By 
drawing  FD'  \  to  b  and  E' D  \  to  <;.] 

632.  Any  two  straights  symmetrical  with  regard  to  an 
angle-bisector  are  called  isogonals  with  reference  to  that  angle. 

633.  If  from  two  points,  one  on  each  of  two  isogonals  with 
respect  to  a  given  angle,  perpendiculars  be  drawn  to  its  arms, 
then, 

I.  The  rectangle  of  the  perpendiculars  to  one  arm  equals 
that  of  those  to  the  other ; 

II.  The  feet  of  the  perpendiculars  are  concyclic ; 

III.  The  join  of  the  feet  of  perpendiculars  from  the 
point  on  either  isogonal  is  perpendicular  to  the  other  isogonal. 

Proof.  By  ~  as, 

GM/GA  =  KQ/KA  ; 

AG/GP  =  AK/KN\ 

.'.GM/GP=  KQ/KN; 

.'.GM.KN=  GP.KQ. 

II.  By  ~  AS, 

F.0..65.  AM/AG=AQ/AK; 

AG/AP  =  AK/AN\ 
.'.  AM/AP  =  AQ/AN; 
.\AM.AN=AP.AQ, 
•.  M,  N,  P,  Q  are  concyclic.  ' 


148  ELEMENTARY  SYNTHETIC  GEOMETRY. 

III.  Since  -4.?,  AMG,  APG  are  r't,  .-.  AMGP is  cyclic; 
.-.  ^  MAG  =  ^  AfPG;     but     ^  G^y^y]/=  ^  A'^Q- 
.-.  ^  MPC;  =  ^  A'^S;     but  also     PG  ±to  AQ; 
.:  PM±  to  AK. 

634.  Inverse.  If  the  rectangle  of  the  JLs  from  2  given  p'ts 
on  one  of  the  arms  of  a  given  ^  equal  the  rectangle  of  the  J_& 
on  the  other  arm,  the  joins  of  the  vertex  and  the  p'ts  are  isog- 
onal  with  respect  to  the  ^ . 

635.  If  3  st's  through  the  vertices  of  a  A  concur,  so  do 
their  isogonals. 

'^  Proof.  Let  AG,  BG  be  the  isogonals 

^M       of  AK,  BK.     Then  by  633, 

AA'  =  /s  A'  > 
A  A'  =P%P'%\ 
hence  /,//  —  P^Px^ 

/.  by  634,  GO  and  KG  are  isogonals.    • 

636.  Two  points  so  related  to  a  triangle  that  the  three 
joins  of  one  to  the  vertices  are  isogonal  to  the  joins  of  the 
other,  are  called  isogonal  conjugates. 

637.  Theorem.  The  six  _L  projections  of  2  isogonal  conju- 
gates on  the  sides  of  the  triangle  are  concyclic  •  and  the  center 
of  this  circle  bisects  their  join. 

Proof.  By  ~  AS, 

BQ'/BM  =  BK/BG  =  BN/BF ; 
.'.BM.BN=^BP'.BQ'', 

.'.  M,  N,  P',  Q!  are  concyclic. 

And  as  the  center  of  ©  round  them  lies  in  the  r't  bi's  of 
MA^and  Q P\  it  is  the  mid  point  of  GK.  Similarly,  Q,  Pare 
on  the  same  ©. 


( 


ANTI-PARALLELS,    ISOGONALS,    SYMMEDIANS.  149 

638.  Corollary  I.  The  circumcenter  and  orthocenter  of  a 
A  are  isog'  conj's;  .'.  a  O  passes  through  A\  B',  C,  D,  E,  F, 
and  the  mid  points  of  AH,  BH,  CH ,  with  center,  TV,  bisecting 
OH,  and  diameter  R. 

This  is  called  the  nine-point  circle. 

Its  center  is  called  the  triangle's  medio-center. 

639.  Corollary.  NI  =  ^R  -  r. 

,*.  the  ninepoint-O   touches  the  in-O  and  each  ex-O  (Feuer- 
bach's  Theorem). 

640.  The  isogonal  conjugate  to  the  centroid  of  a  A  is 
called  the  Lemoinc  point  of  the  A . 

641.  The  isogonals  to  the  medians  of  a  A  are  called  its 
sytnmedians. 

642.  Since  a  median  bisects  all  |Is  to  its  side  of  the  A,  .•. 
by  symmetry  its  symmedian  bisects  all  anti-||s  to  the  side. 

643.  The  Lemoine  point  bisects  3  anti-||s,  which  are  equal, 
since  the  two  halves  going  to  a  side  make  with  it  ^  s  each  =r 
to  the  opposite  ^  of  the  A.  Thus  the  ends  of  any  2  of  these 
are  vertices  of  a  rectangle. 

644.  The  circle  through  the  6  p'ts  in  which  anti-|jS  to  the 
sides  of  a  A,  through  its  Lemoine  p't,  meet  the  sides,  is  called 
the  2d  Lemoine  O  of  that  A. 

645.  Since  the  sides  of  a  A  and  of  its  orthocentric  A  are 
anti-  II  ,  .'.  the  sides  of  the  orthocentric  are  bisected  by  the 
symmedians.  .'.  join  each  angle  of  a  a  to  the  mid  point  of 
that  side  of  the  orthocentric  which  ends  in  its  arms  ;  the  joins 
concur  in  the  Lemoine  p't. 

646.  The  J_s  from  the  Lemoine  p't  to  the  sides  of  a  a  are 
proportional  to  the  sides. 

From  B'  bisecting  b,  and  K,  the  Lemoine  p't,  draw  J_s. 
Then  by~  as,  B' P,  .  KP,' =  B'P,  .  KP,' .  But  since  a 
ABB'  =  A  B'BC,  .-.  B'P,  .  a  =  B'P^  .  c. 

.'.  KP,'/a  =  KP:/c. 


150  ELEMENTARY  SYNTHETIC  GEOMETRY. 

647.  \{  k^,  k^,  k^  are  _Ls  from  K  on  a,  b,  c,  then 


a 


A 


a'^b''  +  c'' 


648.  [Grebe.]  Describe  sq's  APQB,  BUVC,  CXYA  on  the 
sides  of  A  y^^^'  [all  externally  or  all  internally],  and  let  QP^ 
^Fmeet  in  a  ;  PQ,  VU  m  yS ;  UV,  YX  in  y ;  then  aA,  fiB, 
yC  concur  in  K. 

649.  [Mathieu.]  The  Lemoine  point  of  a  triangle  is  the 
center  of  perspective  of  that  triangle  and  its  polar  triangle 
with  respect  to  any  circle. 

Proof.  With  respect  to  circum-O  of  A  ABC,  the  pole  of 
BC  is  P,  of  CA  is  Q,  of  AB  is  R.  LPM  \  to  QR  is  anti-  \  to 
BC.  .'.  PL  =  PB  =PC  =  PM;  .'.  AP  is  a  symmedian  of  a 
ABC,  .'.  the  Lemoine  point  is  in  AP. 


In  same  way,  it  is  in  BQ  and  CR  ;  .*.  it  is  A C of  As  ABC, 
PQR;  and  .•.  of  a  ABC  and  any  other  of  its  polar  As. 

650.  The  joins  of  the  points  of  contact  of  the  in-O  of  a  a 
with  its  opposite  vertices  concur  in  the  Lemoine  p't  of  the  A 
formed  by  joining  the  points  of  contact. 

This  is  called  the  Gergonne  point  of  the  first  A . 

651.  [Schlomilch.]  The  three  joins  of  the  mid  point  of  each 
side  of  a  triangle,  to  the  mid  point  of  the  corresponding 
altitude,  concur  in  the  Lemoine  point. 


ANTI-PARALLELS,   ISOGONALS,    SYMMEDIANS. 


\\\ 


In  A  ABC  let  a,  a'  be  p'ts  in  BC',  ft,  ft'  in  CA,  y,  y'  in 
AB ;  such  that  oiKft',  ftKy',  yKa'  are  the  respective  anti-  jjs 
through  K  [the  Lemoine  p't],  to  AB,  BC,  CA. 

:.  K\^  the  mid  p't  of  these  sects  and  aftft'y'  is  a  rect'. 

The  median  BB'  [of  a  ABC^  cuts  ay'  in  its  mid  p't  J/; 
and  if  MK  meets  ^C  in  iV^,  A'J/  =  KNy^nd  MKN  is  ||  to  aft, 
and  .-.  to  ^^,  the  alt'  from  B.  :.  B'K  is  median  of  a  B'BE 
and  .'.  bisects  BE.   ■ 

652.  [R.  F.  Davis.]  If  of  six  points  a  pair  is  in  each  side 
of  a  triangle  and  concyclic  with  each  other  such  pair,  then  the 
six  are  concyclic. 

Proof.  Ay,  .  Ay,  =  Aft,.  Aft, ;  .-.  A 
is  on  the  radical  axis  of  ©s  a,a,y,y, , 
o(,ot^ft,ft,. 

But  if  these  circles  are  distinct,  yet 
intersect,  their  radical  axis  contains 
their  common  chord  a,a, . 

.".  They  coincide. 


Fig.   270. 


653.  If  a,ft,  anti-||  to  a,ft,  ,  and 

a^y^  anti-ll  to  a,y, ,  and 

ft,y,  anti-||  to  ft,y,  ,  then  the  six  points 

^if  '^^y  fti,  ft,,  Vi,  7,  are  concyclic. 

654.  [Tucker.]   The  six  ends  of  three  equal  anti-parallels 
in  a  triangle  are  concyclic. 


152 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


J I 


Nr- 


V 

L 
Fig.  271. 

Proof.  Let  /'  be  the  in-center  of  A  LMN. 

Since  E' F,  F'D  make  ^  E^FA  =  4  BF'D,  .-.  aFNF' 
is  •]• ;  .-.  a  bisector  of  ^  iV  is  r't  bi'  of  FF^ ;  .\  the  r't  bi'  of 
FF'  passes  through  /'.  Similarly,  the  r't  bi'  of  DD'  passes 
through  /'.  But  since  E'F=  D'E,  :.  D'F  \  to  AC,.:  D\  F, 
F' ,  D  are  concyclic ;  .•.  /'  is  the  center  of  a  O  through  D' ,  F, 
F',  D.     Similarly,  /'  is  center  of  O  through  E' ,  D,  D',  E. 

655.  The  circles  got  by  varying  the  size  of  the  three  equal 
anti-parallels  are  all  called  Tucker's  circles. 

656.  Corollary.  K  is  cost'  with  O  and  O'  the  circumcenters 
of  aABC  ?ind  Aa^y;  and/'  bisects 
the  join  00'.  For,  since  aE'AF  is 
a  II  g'm,  .*.  K  is  /^  C  oi  A  ABC  and 
Aafty. 

Also  OC,  JL  to  tan'  at  C,  is  _L  to 

i?'/;anti-||  to  ^;  and  .'.  O'y,  ||  to  OC, 

is  ±  to  D'E',  .'.  through  J/ (mid  p't 

P'°-'7«.  of    yC    and    of    BE)   a  ||  to    OC 

will    bisect    00'  in    T  and   contain   /'.     Similarly   for  E'F', 

.'.  TisI'. 


ANTI-PARALLELS,   ISOGONALS,    SYMMEDIANS.  1 53 

657.  Straights  through  the  Lemoine  point  of  a  triangle 
parallel  to  its  sides,  are  called  the  Lemoine  parallels  of  that 
triangle. 

658.  The  crosses  of  the  Lemoine  parallels  and  the  sides  of 
a  triangle  are  concyclic. 

For  K  being  now  the  common  vertex  of  3  ||g'ms,  its  joins 
to  the  vertices  of  the  a  bisect  the  other  3  diagonals,  joins  of 
these  crosses,  which  are  thus  3  anti-||^  and  all  equal,  being 
non-|j  sides  of  3  symtras. 

659.  This  Tucker's  circle  through  the  ends  of  the  Lemoine     "N 
parallels  is  called  the  First  Lemoine  Circle. 

660.  The  center  of  the  First  Lemoine  Circle  is  the  mid 
point  of  the  sect  KO.  For  A  a-/?;/ has  become  the  point  A";  so 
/',  bisecting  OO' ,  now  bisects  OK. 

660  (b).  [H.  M.  Taylor.]  The  six  J.  projections  of  the  feet 
of  its  altitudes  on  the  other  sides  of  a  A  are  on  a  Tucker's  0, 
x:alled  its  Taylor's  O. 


^lUf  ^^   h^5>  '^5^? 


< 


CHAPTER   II. 


THE   BROCARD    POINTS. 


66i.  Problem.  To  draw  a  circle  which  shall  pass  through 
a  given  point  and  touch  a  given  straight  at  a  given  point. 

Construction.  The  J_  to  the  st'  at  the  given  p't  and  the  r't 
bi'  of  the  join  of  the  two  given  points  cross  in  the  center. 

662.  Problem.  To  find  within  a  a  a  p't  which  its  sides  will 
contain  after  equal  positive  rotations  about  their  vertices. 

Construction.   Describe  a   ©  passing  through   one  vertex,, 
as  C,  and  touching  the  opposite  side  c 
at  the  next  counter-clockwise  vertex,  A. 
Draw  the  chord  AP  \  to  BC, 
Join  BP,  cutting  the  O  in/2. 
Proof.  Then   [periphery  ^^  on    the 
same  arc  CLA^    ^    AC^  —  -2^  BAD.  — 
•4.  APn  =  4  CBa  [its  alternate  ^]. 
Determination.  Only  one  solution. 

663.  This  p't  fL  is  called  the  positive 
Brocard  p't.  Its  isogonal  conjugate,  the 
negative  Brocard  p't  fl' ,  is  given  by  substi- 
tuting negative  for  positive  in  the  preceding 
problem. 

\  '  664.  D.  and/2'  are  isogonal  conjugates. 

665.  The  magnitude  of  the  angle  of  rotation  for  £1  and  for 
/2'  is  designated  by  00,  and  ^  -j-  co  is  called  the  Brocard  ^  of 
the  A. 

154 


Fig.  273. 


THE  BROCARD  POINTS.  I  55 

666.  The  3  O®  each  passing  through  a  vertex  and  touching 
the  opposite  side  at  the  next  counter-clockwise  vertex  concur 
in  the  positive  Brocard  p't. 

[If  next  clockwise  vertex,  in  /^'.] 

667.  At  A  draw  AX  \  to  BC. 
At  C  make  ^  XCA  =  ^  CBA. 

Then  ^  CBX  is  the  Brocard  ^  of  A  ABC. 

668.  From  ^y,  the  Brocard  ^ ,  ^  -|-  &?,  is  the  same  for  all 
~  As. 

669.  In  the  construction  662  we  may  keep  ^  B  constant 
and  increase  :^  &?,  by  sliding  BC  \  to  itself  until  it  touches 
the  O; 

.-.  if  one  ^  of  a  A  is  fixed,  co  is  greatest  when  the  A  is  •!• ; 
.'.  an  equilateral  A  has  the  greatest  of  all  Brocard  angles, 
which  is  ^  r't  ^ . 

670.  The  arcs  A^IC,  AflB,  BflA,  Bft'C,  CflB,  Cfl' A  are 
called  the  Brocard  arcs  of  the  triangle. 

671.  [Brocard.]  If  O  is  circumcenter  and  K  Lemoine  of 
A  ABC;  and  if  O  on  diameter  OK  cuts  the  Lemoine  jj*  to 
BCt  CA,  AB,  in  a,  /?,  y,  respectively  ;  then 

I.  Ay,  Bar,  C/3  concur  on  the  O;  and  their  cross  is  the 
positive  Brocard  point  D,. 

II.  A^,  By,  Coc  concur  on  the  circle,  and  their  cross  is  the 
negative  Brocard  point  /2'. 

Proof.  Let  £KaF\  FK/3D',  DyKE'  be  the  Lemoine  f. 

Then  ^  OaK  is  r't. 

.-.  aO  _L  to  BC  bisects  BC  in  A'  [since  O  is  circumcenter]. 

,In  same  way  /5(9,  yO  meet  b,  c  in  their  mid  points  B' ,  C. 

Let  Ba,  Cft  cross  in  £1\ 

then  aA'/BA'  =  twice  i.  from  K  on  BC/BC', 

=  twice  i.  from  X  on  d/d  [646] ; 

=r  /3ByCB\ 
.'.   A  aBA'  -^  A/3CB'. 
.-.    ^  BaA'  =  ^  C/3B'  =  ^  0/3n. 


\ 


156 


ELEMENTARY  SYNTHETIC  GEOMETRY. 


.'.  £1  is  concyclic  with  a,  fS,  O,  and  .•.  also  with  y. 
Similarly,  Ay,  Ba  concur  on  this  ©; 
i.e.,  Ay,  Ba,  C^  concur  in  £1  on  the  ©  whose  diam'  is  OK* 


Similarly,  AjS,  By,  Cot  concur  in  a  p't  DJ  on  the  O. 
Moreover,  :^nBC  =  ^nCA  =  :fnAB[{rom'^  AsaBA', 
§CB' ,  yAC'\     .'.  O  is  the  positive  Brocard  p't. 
Similarly,  /2'  is  the  negative  Brocard  p't. 

672.  The  O  on  the  joint  of  the  circumcenter  and  Lemoine 
p't  of  a  A  as  diameter  is  called  the  Brocard  Q  of  the  A ,  from 
the  name  of  its  discoverer. 

673.  The  A  whose  vertices  are  the  _L  projections  of  the 
circumcenter  on  the  Lemoine  ||s  is  called  Brocard's  first  a. 

674.  The  A  whose  vertices  are  the  X  projections  of  the 
circumcenter  on  the  symmedians  is  called  Brocard's  second  A. 

675.  In  A  ABC  to  inscribe  a  A  ~  to  a  given  a. 

In  AB  take  any  p't  D,  and  draw  any  sect  JDF  to  another 
side,  as  AC;  and  at  the  points  D  and  i^  make  ^^  FJDE,  DFE 
equal  to  2  of  the  ^*  of  the  given  a  ;  .-.  ^  ^5"  =  3**  angle. 

Join  AE  and  produce  it  to  meet  the  side  a\n  G  ',  from  G 
draw  GH,  GI,  respectively  |1  to  ED,  EF. 


THE  BROCARD  POINTS  l^T 

Join  HI.      A  HIE  is  the  required  A.       ',       Jl  H  J-  $  • 
For  ^  E  =  7f  G ;  ■     ^ 

also  BE  :  HG  ::AE:  AG  ::EF:GI; 
.-.   A  HGI'^  A  BEF. 

676.  A  ABC<^  A  A'B'C. 

677.  Corollary.  The  sect  from  any  angle  of  a  a  to  //  is. 
twice  the  _L  from  O  on  opposite  side, 

678.  Theorem.  Of  any  a,  O,  >^C,  and  //are  co-st'. 
Proof.  Join  OH,  meeting  A  A'  in  G. 

Bisect  AG  in  X  and  AH  in  V. 

XV  \\  to  and  =  ^GH,  .:  ^  AXY  =  ^  AGH  =  ^  OG^ 
and^F=  OA'.  \_e77\ 
A\so  AD  W  OA'; 
.\XY=  OG  =  ^GH, 
and  AX  =  XG  =  GA'. 
.'.  G  is  ■'^C. 

CO-SYMMEDIAN   AND    CO-BROCARDAL  TRIANGLES. 

679.  Let  the  symmedians  A,K,  B^K,  C,K  of  a  A^B.C,  be 
produced  to  meet  the  circum-O  in  A^,  B^,  C^,  and  let  the  op- 
posite sides  j?,^', , -5^^,  of  the  quad' i?, 6^,^52 Cj  be  produced  to 
meet  in  L. 

Then  the  polar  of  L  passes  through  K,  the  cross  of  B^B^ . 

But  since  A^A^  is  a  symmedian  of  the  a  A^B^C,,  the  tan- 
gents at  B^  and  C^  cross  on  A^KA^  produced,  .•.  A^KA^  is  the 
polar  of  L,  and  the  tangents  at  B^ ,  C^  must  cross  on  this  st , 
which  is  consequently  a  symmedian  of  the  a  A^BC^  also  ; 
similarly  for  the  straights  B,KB^ ,  C^KC^. 

These  as  A^B^C^,  A,B^C,  having  the  same  symmedians  are 
called  co-symmedian  triangles. 

They  have  the  same  Lemoine  p't,  and  the  same  circum- 
center,  consequently  the  same  Brocard    ©.      Also  the  same 


158  ELEMENTARY  SYNTHETIC  GEOMETRY. 

Brocard  ^  ,  and  the  same  Brocard  p'ts,  the  same  first 
Lemoine  ©  ,  and  the  same  second  Lemoine  O ,  also  the 
Tucker  Os  of  one  are  Tucker  Qs  of  the  other,  though  a  par- 
ticular Tucker  O  of  one  is  not  always  the  same  Tucker  O  of 
the  other.  Thus  the  Taylor  Q  of  one  is  not  the  Taylor  ©  of 
the  other. 

680.  If  2  As  are  co-symmedian,  the  sides  of  one  are  propor- 
tional to  the  medians  of  the  other. 

For  ^s  C,A^B,  =  C,A,A,  +  B.A^A,  =  KC,A,  +  KB,A,  = 
GCjBj  -[-  GB^C^,  since  G  and  AT  are  isogonal  conjugates. 

Hence  QA,B,  =  BfiC;  =  A/C/'G,  where  C/'  is  the  cross 
of  C,G  with  a  i  to  B,G  through  A/ .  Similarly,  ^  A,B,C,  = 
i-C/'GA/. 

Thus,  A  A^B^C^  ~  A  C^"GA^' ,  each  of  whose  sides  is  -^ 
the  corresponding  median  of  a  A^B^C^. 

681.  To  show  that  any  A  y^j^jC,  has  corresponding  to  it  not 
only  the  co-syn^edian  a  A^B^C^,  but  an  infinity  of  others  having 
the  same  Brocard  p'ts,  Lemoine  p't,  Brocard  ©,  ist  Lemoine 
0,  2d  Lemoine  ©,  we  should  study  the  triangle's  Brocard 
ellipse,  but  this  would  carry  us  beyond  strictly  elementary 
geometry. 

A  system  of  As  thus  connected  are  called  co-Brocardal 
AS. 

682.  If  a  a  A,B,C,  be  inscribed  m  a  given  a  ABC,  the  ©s 
AB,C,  ,  BA,C, ,  CA,B,  concur.     For  let  ©s  AB,C, ,  BA,C,  meet 

in  O. 

Then   since  B,OC,  =  st'  ^-A,  and  C,OA,  =  sV   ^-B, 

we  have     B,OA,  =  2  st'  ^s  -  [sf  ^  -  A]  -  [sf  ^  - B] 
=  A-^B  =  st'  4  -C; 

.'.  the  quad'  A,OB,C  is  cyclic. 

683.  If  the  A  A.B^C,  inscribed  in  A  ABC  is  ~  and  co-sensal 
to  it,  and  A,  falls  on  a,  then 

B0C  =  A-\~A,  =  2A; 


THE  BROCARD  POINTS.  I59 

simKarly,  CO  A  =28,     and     A0B  =  2C\ 

therefore  0  is  the  circumcenter. 

684.  If  A^  falls  on  c,  the  Os  concur  in  D,. 

If  A^  falls  on  d,  the  Os  concur  in  fi'. 

In  the  first  case,  the  a  and  its  inscribed   A  have  the  same 
positive  Brocard  p't. 

In  the  second  cdse,  the  same  negative  Brocard  p't. 

685.  Similar  as  have  the  same  Brocard  ^  a?. 

686.  O  is  the  circumcenter  of  A  ABC;  OA^ ,  OB^ ,  OC,  are 
drawn  to  a,  b,  c,  so  that   ^OA^A  =  yf  OB^B  =   ^  OC,C. 

Show  that  A  A^B^C^  is  ~  and  co-sensal  to  a  ABC. 
Show  also  that  O  is  the  orthocenter  of  A  A^B^C, . 


I60  ELEMENTARY  SYNTHETIC   GEOMETRY. 


EXERCISES   ON   BOOK   VIII. 

1.  Through  the  mid  point  of  each  side  of  a  A  are  drawn  ±8  to  the 
other  2  sides.    Show  that  the  2  as  thus  formed  have  the  same  Lemoine  p't. 

2.  Show  that  the  Lemoine  p't  of  a  A  is  the  centroid  of  its  1  projec- 
tions on  the  sides  ;  and  inversely. 

3.  Find  a  p't  within  a  A  such  that  the  sum  of  the  squares  of  its  Xs  on 
the  sides  is  a  minimum. 

[The  p't  must  be  the  centroid  of  its  1  projections  on  the  sides,  and 
.*.  the  Lemoine  p't.] 

4.  The  joins  of  the  circumcenter  of  a  A  to  its  vertices  are  J.  to  the 
sides  of  its  orthocentric  A. 

5.  Brocard's  first  A  is  in  perspective  with  its  original  A. 

6.  Brocard's  first  A  is  ~  but  not  cosensal  to  its  original. 

7.  The  join  of  the  circumcenter  and  Lemoine  p't  of  a  A  is  the  r't  bi' 
of  the  join  of  its  Brocard  p'ts. 

[  ^  aaK  =  ^  ChBC  =00  =  a'CB  =  n'a/C] 

8.  If  T  is  the  center  of  Brocard's  O,  then  ^  flTfl'  =  2ilTK  =  4cj. 

9.  If  AK,  BK,  CK  meet  the  Brocard  ©  in  a',  /J',  y' ,  then  a'  is  the 
cross  of  the  Brocard  arcs  AilC,  Afl'B  ;  ff  is  the  cross  of  BJCIA,  BLl'C; 
and  y'  is  the  cross  of  CD.B,  Cfl'A. 

10.  [Dewulf.J  If  through  the  Brocard  p't  £1  three  0»  be  described 
each  passing  through  two  vertices  of  a  ABC,  the  A  formed  by  their 
centers  has  the  circumcenter  of  ABC  for  one  of  its  Brocard  p'ts. 

11.  The  join  of  any  two  p'ts,  and  the  join  of  their  isogonal  conjugates, 
with  respect  to  a  A,  subtend  at  any  vertex  of  the  a  ^  either  =  or  sup- 
plemental. 

12.  If  three  st's  through  the  v^nices  of  a  A  meet  the  opposite  sides 
co-st'ly,  so  do  their  isogonals. 

13.  The  joins  of  the  ±  projections  of  the  Lemoine  p't  on  the  sides  of 
the  A  are  X  to  the  medians. 

14.  If  on  a  given  sect,  and  on  the  same  side  of  it,  be  described  six. 
A^  ~  to  a  given  A,  the  vertices  are  concyclic. 

15.  In  Fig.  275,  Aa,  Bfi,  Cy  concur. 


INDEX. 


(The  numbers  refer  to  .he  pages.) 


Acute 13 

Adjacent   7 

Adrian ' 108 

Altitude 43 

Angle 5 

Angle,  Brocard 154 

inscribed 40 

periphery 40 

tanchord 29 

Angles,  alternate 40 

Anti-parallels 146 

Apollonius Ill 

Archimedes 108 

Area 104 

Arms  of  angle 6 

Axis  of  perspective 119 

of  symmetry 18 

radical 142 

Base 86 

Bearer I2I 

Bi-radial    5 

Bisectors  of  angles 34 

Brianchon 140 

Brocard  angle 154 

arcs 155 

circle I55 

points 154 

first  triangle 156 

second  triangle    156 


Broken  line 37 

Cenquad 71 

Center 8 

Center  of  inversion 129 

of  involution 141 

of  perspective 95 

of  similitude • .  -  95 

Centimeter 104 

Centroid    98 

Ceva 118 

Chord 10 

Circle S 

Circle,  Apollonius 1 1 1 

Brocard i55 

circum 34 

ex 35 

in 34 

Lemoine's,  1st 153 

2d 149 

Nine-point 149 

Taylor i53 

Circles,  Tucker's 152 

Circular  measure 108 

Circum-center 34 

-circle 34 

-radius 34 

Clockwise 5 

Co-Brocardal.triangles I57 

Complement  of  angle 13 

161 


l62 


INDEX. 


Complete  quadrilateral 126 

Comquad 126 

Conclusion g 

Concurrent 34 

Concyclic , 41 

Congruent 3 

Conjugate  points   137 

Conjugates,  harmonic 121 

isogonal 148 

Construction 9 

Contraparallelogram 132 

Convex 38 

Corresponding  points 95 

Co-sensal 146 

Co-straight   31 

Cosymmedian  triangles 157 

Cross 123 

Cross-ratio 137 

Curve 4 

Cyclic 41 

polygon 70 

Davis 151 

Deltoid 49 

Determination 9 

Diagonal 38 

points 126 

triangle 126 

Diameter 10 

Dual ....   124 

Egyptians 108 

Element 123 

Equal 6 

Equi-cross 140 

Equivalent 85 

Escribed  circle 35 

Ex-center 35 

Ex-circle 35 

Explement 7 

Explemental  arcs 12 

Exterior  angle ". ..     35 

Euler 108 

Feuerbach  149 


Figures 3 

reciprocal 124 

Fraction ....     92 

G-Iine 60 

Gergonne 150 

Grebe 150 

Harmonic  conjugates 121 

pencil 122 

progression 121 

range 97 

Harmonically  divided 97 

Hart , 132 

Hemiplanes 5 

Heron 109 

Hexagon 38 

Brianchon's 140 

Pascal's 140 

Hexagram 141 

Hindoos 108 

Hypothenuse 35 

Hypothesis 9 

In-center 34 

-circle <3^  '\^ 

-radius 34 

Incommensurable 92 

Inscribed  angle 40 

Inverse  points 129 

Inversion 129 

Involution 141 

Isogonal  conjugates 148 

Isogonals 146 

Isosceles 35 

Join 123 

K  in  Lemoine  point 

Lambert 108 

Lemoine's  first  circle 153 

second  circle 149 

parallels..., 153 

point 149 

Length 104 

Lindemann 108 

Line 2 


INDEX. 


163 


Line,  Simson 119 

Linkage 130 

Ludolf. 108 

Lune 62 

Major 13 

Maximum 142 

Measurement 104 

Median    43 

Medio-center 149 

Menelaus 117 

Milne 142 

Minimum 142 

Minor , 13 

Multiple 92 

Nine-point  circle 149 

Oblique 24 

Obtuse 13 

Opposite  points 60 

Orthocenter 57 

Orthocentric  triangle 147 

Parallelogram 46 

Parallels '. 30 

Pascal 140 

Peaucellier 130 

Pencil 121 

Perigon 6 

Perimeter 37 

Periphery  angle 40 

Perspective 95 

Perspective-center 121 

Pi  (tt) 108 

Plane 3 

Point 2 

of  Gergonne 150 

Lemoine 149 

Points,  Brocard 154 

conjugate 121 

corresponding 95 

double 141 

Polar 75,  133 

Polar  reciprocal 136 

Polar  triangle 78 


Polar  polygon 79 

polarization 136 

Pole  of  circle 62 

Pole  and  Polar 133 

Polygon 37 

convex 30 

star 33 

undivided 37 

Problem 9 

Projection  1 119 

Projection-axis 11 1 

Projection-straights 96 

Projective 95 

Proportion 92 

Ptolemy 102 

q- radius 67 

q-pole 67 

Quadrangle 126 

Quadrant 13 

Quadrilateral 33 

complete 126 

Radian 109 

Radical  axis 142 

Radius 8 

Radius  of  inversion 129. 

Range 121 

harmonic 97 

Ratio 92 

cross 137 

Rays 5 

Reciprocal 124 

Rectangle 52 

Reflex 13 

Regular 38 

Revolution 4 

Rhombus 52 

Rotation 5 

-centre 57 

Schlomilch , 150 

Secant 22 

Sect 4 

Segments 97 


164 


INDEX. 


Sense 5 

Similarity 95 

Similitude,  ratio  of 95 

Simson IIO 

Sphere .• 60 

Spherical  excess 88 

Square 53 

Straight 4 

angle 6 

Submultiple 92 

Sum 7 

Summit 126 

Supplement 7 

Surface I 

Symmedians 146 

Symcenter 19 

Symcentry 19 

Symmetry 18 

Symtra 49 

Tanchord  angle 40 

Tangent 22 

Tangent  circles   28 

Theorem 9 

Bordage 84 

Brahmegupla 84 

Brianchon 140 


Theorem  of  Brocard 157 

Ceva 118 

Davis 151 

Dewulf 160 

Feuerbach i^g 

Grebe 150 

Hart 132 

Mathieu 150 

Menelaus 117 

Pappus 90 

Pascal 140 

Ptolemy 102 

Schlomilch..    150 

Tucker 151 

Trace , 

Transversal 29 

Trapezoid 91 

Triangle,  Brocard's  first 157 

second. . .     ...  157 

diagonal 126 

orthocentric 147 

self-conjugate 134 

Tucker's  circles > . . .   152 

Unit 104 

Vertex 121 


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